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#6: Post edited by

Lundin‭ · 2020-08-21T13:13:06Z (about 4 years ago)
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Yes, the second line invokes undefined behavior.
First of all, according to C17 6.5.2.1 regarding array subscripting, an expression `E1[E2]` is just "syntactic sugar" for `*((E1)+(E2)))`. So what applies here is actually the binary + operator. More info regarding why `[]` is actually never used with an array operand [here](https://stackoverflow.com/questions/55747822/do-pointers-support-array-style-indexing).
~~(The code `char *ptr = arr[-1];` isn't correct C. That's another story, so I'm assuming this is a typo and that the intention was `char *ptr = &arr[-1];`)~~
So your example is equivalent to `char *ptr = *((arr) + (-1));`, where `arr` "decays" into a pointer to the first element. The `arr` operand ends up as a pointer type and the `-1` operand is an integer type.
C17 6.5.6/8 then provides the following text for additive operators:
> When an expression that has integer type is added to or subtracted from a pointer, the
result has the type of the pointer operand. /--/
> If both the pointer
operand and the result point to elements of the same array object, or one past the last
element of the array object, the evaluation shall not produce an overflow; otherwise, the
behavior is undefined.
In this case, the result does not point inside `arr` so "otherwise, the behavior is undefined".
That is, the evaluation of the + operator is what first causes undefined behavior, before the de-referencing.
---
We can demonstrate this UB with gcc x86 -O3 by first compiling:
int arr[3] = {1,2,3};
printf("%d\n", arr[0]);
which disassembles into
mov edi, offset .L.str
mov esi, 1
That is, as part of some calling convention ESI gets directly loaded with the value 1 , equivalent to what would have been stored in arr[0] if the array got allocated. If I change this to `printf("%d\n", arr[-1]);`, the instruction for setting up ESI is simply removed from the disassembly and I suppose the program prints whatever garbage value that happened to be stored inside ESI. The compiler doesn't even attempt to de-reference the variable by fetching a value from the stack corresponding to memory address `arr - 1`.

Yes, the second line invokes undefined behavior.
First of all, according to C17 6.5.2.1 regarding array subscripting, an expression `E1[E2]` is just "syntactic sugar" for `*((E1)+(E2)))`. So what applies here is actually the binary + operator. More info regarding why `[]` is actually never used with an array operand [here](https://stackoverflow.com/questions/55747822/do-pointers-support-array-style-indexing).
So your example is equivalent to `char *ptr = &*((arr) + (-1));`, where `arr` "decays" into a pointer to the first element. The `arr` operand ends up as a pointer type and the `-1` operand is an integer type.
C17 6.5.6/8 then provides the following text for additive operators:
> When an expression that has integer type is added to or subtracted from a pointer, the
result has the type of the pointer operand. /--/
> If both the pointer
operand and the result point to elements of the same array object, or one past the last
element of the array object, the evaluation shall not produce an overflow; otherwise, the
behavior is undefined.
In this case, the result does not point inside `arr` so "otherwise, the behavior is undefined".
That is, the evaluation of the + operator is what first causes undefined behavior, before the de-referencing.
---
We can demonstrate this UB with gcc x86 -O3 by first compiling:
int arr[3] = {1,2,3};
printf("%d\n", arr[0]);
which disassembles into
mov edi, offset .L.str
mov esi, 1
That is, as part of some calling convention ESI gets directly loaded with the value 1 , equivalent to what would have been stored in arr[0] if the array got allocated. If I change this to `printf("%d\n", arr[-1]);`, the instruction for setting up ESI is simply removed from the disassembly and I suppose the program prints whatever garbage value that happened to be stored inside ESI. The compiler doesn't even attempt to de-reference the variable by fetching a value from the stack corresponding to memory address `arr - 1`.

#5: Post edited by

Lundin‭ · 2020-08-18T06:28:56Z (about 4 years ago)

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Yes, the second line invokes undefined behavior.
First of all, according to C17 6.5.2.1 regarding array subscripting, an expression `E1[E2]` is just "syntactic sugar" for `*((E1)+(E2)))`. So what applies here is actually the binary + operator. More info regarding why `[]` is actually never used with an array operand [here](https://stackoverflow.com/questions/55747822/do-pointers-support-array-style-indexing).
So your example is equivalent to `char *ptr = *((arr) + (-1));`, where `arr` "decays" into a pointer to the first element. The `arr` operand ends up as a pointer type and the `-1` operand is an integer type.
C17 6.5.6/8 then provides the following text for additive operators:
> When an expression that has integer type is added to or subtracted from a pointer, the
result has the type of the pointer operand. /--/
> If both the pointer
operand and the result point to elements of the same array object, or one past the last
element of the array object, the evaluation shall not produce an overflow; otherwise, the
behavior is undefined.
In this case, the result does not point inside `arr` so "otherwise, the behavior is undefined".
That is, the evaluation of the + operator is what first causes undefined behavior, before the de-referencing.
---
We can demonstrate this UB with gcc x86 -O3 by first compiling:
int arr[3] = {1,2,3};
printf("%d\n", arr[0]);
which disassembles into
mov edi, offset .L.str
mov esi, 1
That is, as part of some calling convention ESI gets directly loaded with the value 1 , equivalent to what would have been stored in arr[0] if the array got allocated. If I change this to `printf("%d\n", arr[-1]);`, the instruction for setting up ESI is simply removed from the disassembly and I suppose the program prints whatever garbage value that happened to be stored inside ESI. The compiler doesn't even attempt to de-reference the variable by fetching a value from the stack corresponding to memory address `arr - 1`.

Yes, the second line invokes undefined behavior.
First of all, according to C17 6.5.2.1 regarding array subscripting, an expression `E1[E2]` is just "syntactic sugar" for `*((E1)+(E2)))`. So what applies here is actually the binary + operator. More info regarding why `[]` is actually never used with an array operand [here](https://stackoverflow.com/questions/55747822/do-pointers-support-array-style-indexing).
(The code `char *ptr = arr[-1];` isn't correct C. That's another story, so I'm assuming this is a typo and that the intention was `char *ptr = &arr[-1];`)
So your example is equivalent to `char *ptr = *((arr) + (-1));`, where `arr` "decays" into a pointer to the first element. The `arr` operand ends up as a pointer type and the `-1` operand is an integer type.
C17 6.5.6/8 then provides the following text for additive operators:
> When an expression that has integer type is added to or subtracted from a pointer, the
result has the type of the pointer operand. /--/
> If both the pointer
operand and the result point to elements of the same array object, or one past the last
element of the array object, the evaluation shall not produce an overflow; otherwise, the
behavior is undefined.
In this case, the result does not point inside `arr` so "otherwise, the behavior is undefined".
That is, the evaluation of the + operator is what first causes undefined behavior, before the de-referencing.
---
We can demonstrate this UB with gcc x86 -O3 by first compiling:
int arr[3] = {1,2,3};
printf("%d\n", arr[0]);
which disassembles into
mov edi, offset .L.str
mov esi, 1
That is, as part of some calling convention ESI gets directly loaded with the value 1 , equivalent to what would have been stored in arr[0] if the array got allocated. If I change this to `printf("%d\n", arr[-1]);`, the instruction for setting up ESI is simply removed from the disassembly and I suppose the program prints whatever garbage value that happened to be stored inside ESI. The compiler doesn't even attempt to de-reference the variable by fetching a value from the stack corresponding to memory address `arr - 1`.

#4: Post edited by

Lundin‭ · 2020-08-11T13:17:41Z (about 4 years ago)

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Yes, the second line invokes undefined behavior.
First of all, according to C17 6.5.2.1 regarding array subscripting, an expression `E1[E2]` is just "syntactic sugar" for `*((E1)+(E2)))`. So what applies here is actually the binary + operator. More info regarding why `[]` is actually never used with an array operand [here](https://stackoverflow.com/questions/55747822/do-pointers-support-array-style-indexing).
So your example is equivalent to `char *ptr = *((arr) + (-1));`, where `arr` "decays" into a pointer to the first element. The `arr` operand ends up as a pointer type and the `-1` operand is an integer type.
C17 6.5.6/8 then provides the following text for additive operators:
> When an expression that has integer type is added to or subtracted from a pointer, the
result has the type of the pointer operand. /--/
> If both the pointer
operand and the result point to elements of the same array object, or one past the last
element of the array object, the evaluation shall not produce an overflow; otherwise, the
behavior is undefined.
In this case, the result does not point inside `arr` so "otherwise, the behavior is undefined".
That is, the evaluation of the + operator is what first causes undefined behavior, before the de-referencing.
---
We can demonstrate this UB with gcc x86 -O3 by first compiling:
int arr[3] = {1,2,3};
printf("%d\n", arr[0]);
which disassembles into
mov edi, offset .L.str
mov esi, 1
~~That is, as part of some calling convention ESI gets directly loaded with the value 1 from arr[0]. If I change this to `printf("%d~~
", arr[-1]);`, the instruction for setting up ESI is simply removed from the disassembly and I suppose the program prints whatever garbage value that happened to be stored inside ESI. The compiler doesn't even attempt to de-reference the variable by fetching a value from the stack corresponding to memory address `arr - 1`.

Yes, the second line invokes undefined behavior.
First of all, according to C17 6.5.2.1 regarding array subscripting, an expression `E1[E2]` is just "syntactic sugar" for `*((E1)+(E2)))`. So what applies here is actually the binary + operator. More info regarding why `[]` is actually never used with an array operand [here](https://stackoverflow.com/questions/55747822/do-pointers-support-array-style-indexing).
So your example is equivalent to `char *ptr = *((arr) + (-1));`, where `arr` "decays" into a pointer to the first element. The `arr` operand ends up as a pointer type and the `-1` operand is an integer type.
C17 6.5.6/8 then provides the following text for additive operators:
> When an expression that has integer type is added to or subtracted from a pointer, the
result has the type of the pointer operand. /--/
> If both the pointer
operand and the result point to elements of the same array object, or one past the last
element of the array object, the evaluation shall not produce an overflow; otherwise, the
behavior is undefined.
In this case, the result does not point inside `arr` so "otherwise, the behavior is undefined".
That is, the evaluation of the + operator is what first causes undefined behavior, before the de-referencing.
---
We can demonstrate this UB with gcc x86 -O3 by first compiling:
int arr[3] = {1,2,3};
printf("%d\n", arr[0]);
which disassembles into
mov edi, offset .L.str
mov esi, 1
That is, as part of some calling convention ESI gets directly loaded with the value 1 , equivalent to what would have been stored in arr[0] if the array got allocated. If I change this to `printf("%d
", arr[-1]);`, the instruction for setting up ESI is simply removed from the disassembly and I suppose the program prints whatever garbage value that happened to be stored inside ESI. The compiler doesn't even attempt to de-reference the variable by fetching a value from the stack corresponding to memory address `arr - 1`.

#3: Post edited by

Lundin‭ · 2020-08-11T13:16:48Z (about 4 years ago)

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Yes, the second line invokes undefined behavior.
First of all, according to C17 6.5.2.1 regarding array subscripting, an expression `E1[E2]` is just "syntactic sugar" for `*((E1)+(E2)))`. So what applies here is actually the binary + operator. More info regarding why `[]` is actually never used with an array operand [here](https://stackoverflow.com/questions/55747822/do-pointers-support-array-style-indexing).
So your example is equivalent to `char *ptr = *((arr) + (-1));`, where `arr` "decays" into a pointer to the first element. The `arr` operand ends up as a pointer type and the `-1` operand is an integer type.
C17 6.5.6/8 then provides the following text for additive operators:
> When an expression that has integer type is added to or subtracted from a pointer, the
result has the type of the pointer operand. /--/
> If both the pointer
operand and the result point to elements of the same array object, or one past the last
element of the array object, the evaluation shall not produce an overflow; otherwise, the
behavior is undefined.
In this case, the result does not point inside `arr` so "otherwise, the behavior is undefined".
That is, the evaluation of the + operator is what first causes undefined behavior, before the de-referencing.
---
We can demonstrate this UB with gcc x86 -O3 by first compiling:
int arr[3] = {1,2,3};
printf("%d\n", arr[0]);
which disassembles into
mov edi, offset .L.str
mov esi, 1
~~That is, as part of some calling convention ESI gets directly loaded with the value 1 from arr[0]. If I change this to `printf("%d~~
", arr[-1]);`, the instruction for setting up ESI is simply removed from the disassembly and I suppose the program prints whatever garbage value that happened to be stored inside ESI. The compiler doesn't even attempt to de-reference the variable by fetching a value from the stack corresponding to memory address `arr - 1`. The array was never de-referenced since the whole array got optimized away.

Yes, the second line invokes undefined behavior.
First of all, according to C17 6.5.2.1 regarding array subscripting, an expression `E1[E2]` is just "syntactic sugar" for `*((E1)+(E2)))`. So what applies here is actually the binary + operator. More info regarding why `[]` is actually never used with an array operand [here](https://stackoverflow.com/questions/55747822/do-pointers-support-array-style-indexing).
So your example is equivalent to `char *ptr = *((arr) + (-1));`, where `arr` "decays" into a pointer to the first element. The `arr` operand ends up as a pointer type and the `-1` operand is an integer type.
C17 6.5.6/8 then provides the following text for additive operators:
> When an expression that has integer type is added to or subtracted from a pointer, the
result has the type of the pointer operand. /--/
> If both the pointer
operand and the result point to elements of the same array object, or one past the last
element of the array object, the evaluation shall not produce an overflow; otherwise, the
behavior is undefined.
In this case, the result does not point inside `arr` so "otherwise, the behavior is undefined".
That is, the evaluation of the + operator is what first causes undefined behavior, before the de-referencing.
---
We can demonstrate this UB with gcc x86 -O3 by first compiling:
int arr[3] = {1,2,3};
printf("%d\n", arr[0]);
which disassembles into
mov edi, offset .L.str
mov esi, 1
That is, as part of some calling convention ESI gets directly loaded with the value 1 from arr[0]. If I change this to `printf("%d
", arr[-1]);`, the instruction for setting up ESI is simply removed from the disassembly and I suppose the program prints whatever garbage value that happened to be stored inside ESI. The compiler doesn't even attempt to de-reference the variable by fetching a value from the stack corresponding to memory address `arr - 1`.

#2: Post edited by

Lundin‭ · 2020-08-11T13:16:24Z (about 4 years ago)

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Yes, the second line invokes undefined behavior.
First of all, according to C17 6.5.2.1 regarding array subscripting, an expression `E1[E2]` is just "syntactic sugar" for `*((E1)+(E2)))`. So what applies here is actually the binary + operator. More info regarding why `[]` is actually never used with an array operand [here](https://stackoverflow.com/questions/55747822/do-pointers-support-array-style-indexing).
So your example is equivalent to `char *ptr = *((arr) + (-1));`, where `arr` "decays" into a pointer to the first element. The `arr` operand ends up as a pointer type and the `-1` operand is an integer type.
C17 6.5.6/8 then provides the following text for additive operators:
> When an expression that has integer type is added to or subtracted from a pointer, the
result has the type of the pointer operand. /--/
> If both the pointer
operand and the result point to elements of the same array object, or one past the last
element of the array object, the evaluation shall not produce an overflow; otherwise, the
behavior is undefined.
In this case, the result does not point inside `arr` so "otherwise, the behavior is undefined".
That is, the evaluation of the + operator is what first causes undefined behavior, before the de-referencing.
---
We can demonstrate this UB with gcc x86 -O3 by first compiling:
int arr[3] = {1,2,3};
printf("%d\n", arr[0]);
which disassembles into
mov edi, offset .L.str
mov esi, 1
~~That is, as part of some calling convention ESI gets directly loaded with the value 1 from arr[0]. If I change this to `printf("%d~~
", arr[-1]);`, the instruction for setting up ESI is simply removed from the disassembly and I suppose the program prints whatever garbage value that happened to be stored inside ESI. The compiler doesn't even attempt to fetch a value from the stack corresponding to memory address `arr - 1`.

Yes, the second line invokes undefined behavior.
First of all, according to C17 6.5.2.1 regarding array subscripting, an expression `E1[E2]` is just "syntactic sugar" for `*((E1)+(E2)))`. So what applies here is actually the binary + operator. More info regarding why `[]` is actually never used with an array operand [here](https://stackoverflow.com/questions/55747822/do-pointers-support-array-style-indexing).
So your example is equivalent to `char *ptr = *((arr) + (-1));`, where `arr` "decays" into a pointer to the first element. The `arr` operand ends up as a pointer type and the `-1` operand is an integer type.
C17 6.5.6/8 then provides the following text for additive operators:
> When an expression that has integer type is added to or subtracted from a pointer, the
result has the type of the pointer operand. /--/
> If both the pointer
operand and the result point to elements of the same array object, or one past the last
element of the array object, the evaluation shall not produce an overflow; otherwise, the
behavior is undefined.
In this case, the result does not point inside `arr` so "otherwise, the behavior is undefined".
That is, the evaluation of the + operator is what first causes undefined behavior, before the de-referencing.
---
We can demonstrate this UB with gcc x86 -O3 by first compiling:
int arr[3] = {1,2,3};
printf("%d\n", arr[0]);
which disassembles into
mov edi, offset .L.str
mov esi, 1
That is, as part of some calling convention ESI gets directly loaded with the value 1 from arr[0]. If I change this to `printf("%d
", arr[-1]);`, the instruction for setting up ESI is simply removed from the disassembly and I suppose the program prints whatever garbage value that happened to be stored inside ESI. The compiler doesn't even attempt to de-reference the variable by fetching a value from the stack corresponding to memory address `arr - 1`. The array was never de-referenced since the whole array got optimized away.

#1: Initial revision by

Lundin‭ · 2020-08-11T13:08:40Z (about 4 years ago)

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Yes, the second line invokes undefined behavior.

First of all, according to C17 6.5.2.1 regarding array subscripting, an expression `E1[E2]` is just "syntactic sugar" for `*((E1)+(E2)))`. So what applies here is actually the binary + operator. More info regarding why `[]` is actually never used with an array operand [here](https://stackoverflow.com/questions/55747822/do-pointers-support-array-style-indexing).

So your example is equivalent to `char *ptr = *((arr) + (-1));`, where `arr` "decays" into a pointer to the first element. The `arr` operand ends up as a pointer type and the `-1` operand is an integer type.

C17 6.5.6/8 then provides the following text for additive operators:

> When an expression that has integer type is added to or subtracted from a pointer, the
result has the type of the pointer operand. /--/  
> If both the pointer
operand and the result point to elements of the same array object, or one past the last
element of the array object, the evaluation shall not produce an overflow; otherwise, the
behavior is undefined.

In this case, the result does not point inside `arr` so "otherwise, the behavior is undefined".

That is, the evaluation of the + operator is what first causes undefined behavior, before the de-referencing.

---

We can demonstrate this UB with gcc x86 -O3 by first compiling:

    int arr[3] = {1,2,3};
    printf("%d\n", arr[0]);

which disassembles into 

        mov     edi, offset .L.str
        mov     esi, 1

That is, as part of some calling convention ESI gets directly loaded with the value 1 from arr[0]. If I change this to `printf("%d\n", arr[-1]);`, the instruction for setting up ESI is simply removed from the disassembly and I suppose the program prints whatever garbage value that happened to be stored inside ESI. The compiler doesn't even attempt to fetch a value from the stack corresponding to memory address `arr - 1`.

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