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Q&A What is the difference between operator precedence and order of evaluation?

When doing something simple such as this int a=1; int b=2; int c=3; printf("%d\n", a + b * c); then I was told that operator precedence guarantees that the code is equivalent to a + (b * c)...

1 answer · posted 4y ago by Lundin‭ · edited 4y ago by Lundin‭

Question c operator-precedence order-of-evaluation unspecified-behavior

#4: Post edited by

Lundin‭ · 2021-02-01T11:54:34Z (almost 4 years ago)
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When doing something simple such as this
int a=1;
int b=2;
int c=3;
printf("%d\n", a + b * c);
~~then I was told that operator precedence guarantees that the code is equivalent to `a + (b * c)`, since `*` has higher precedence than `+`. And so the result is guaranteed to be 7 and not 9.~~
However, when I modify the above example like this:
#include <stdio.h>
int a (void) { printf("%s ",__func__); return 1; }
int b (void) { printf("%s ",__func__); return 2; }
int c (void) { printf("%s ",__func__); return 3; }
int main (void)
{
printf("%d\n", a() + b() * c());
return 0;
}
Then I get the output `a b c 7`. How is this possible?
Shouldn't operator precedence guarantee that `b()` is executed before `a()`?

When doing something simple such as this
int a=1;
int b=2;
int c=3;
printf("%d\n", a + b * c);
then I was told that operator precedence guarantees that the code is equivalent to
`a + (b * c)`, since `*` has higher precedence than `+`. And so the result is guaranteed to be 7 and not 9.
However, when I modify the above example like this:
#include <stdio.h>
int a (void) { printf("%s ",__func__); return 1; }
int b (void) { printf("%s ",__func__); return 2; }
int c (void) { printf("%s ",__func__); return 3; }
int main (void)
{
printf("%d\n", a() + b() * c());
return 0;
}
Then I get the output `a b c 7`. How is this possible?
Shouldn't operator precedence guarantee that `b()` is executed before `a()`?

#3: Post edited by

Lundin‭ · 2020-09-24T06:41:34Z (about 4 years ago)

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c c++ operator-precedence order-of-evaluation unspecified-behavior

c operator-precedence order-of-evaluation unspecified-behavior

#2: Post edited by

Lundin‭ · 2020-09-23T14:17:28Z (about 4 years ago)

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When doing something simple such as this
int a=1;
int b=2;
int c=3;
printf("%d\n", a + b * c);
then I was told that operator precedence guarantees that the code is equivalent to `a + (b * c)`, since `*` has higher precedence than `+`. And so the result is guaranteed to be 7 and not 9.
However, when I modify the above example like this:
#include <stdio.h>
int a (void) { printf("%s ",__func__); return 1; }
int b (void) { printf("%s ",__func__); return 2; }
int c (void) { printf("%s ",__func__); return 3; }
int main (void)
{
printf("%d\n", a() + b() * c());
return 0;
}
Then I get the output `a b c 7`. How is this possible? And when I try the identical code on another compiler, I get the opposite order `c b a 7`.
Shouldn't operator precedence guarantee that `b()` is executed before `a()`?

When doing something simple such as this
int a=1;
int b=2;
int c=3;
printf("%d\n", a + b * c);
then I was told that operator precedence guarantees that the code is equivalent to `a + (b * c)`, since `*` has higher precedence than `+`. And so the result is guaranteed to be 7 and not 9.
However, when I modify the above example like this:
#include <stdio.h>
int a (void) { printf("%s ",__func__); return 1; }
int b (void) { printf("%s ",__func__); return 2; }
int c (void) { printf("%s ",__func__); return 3; }
int main (void)
{
printf("%d\n", a() + b() * c());
return 0;
}
Then I get the output `a b c 7`. How is this possible?
Shouldn't operator precedence guarantee that `b()` is executed before `a()`?

#1: Initial revision by

Lundin‭ · 2020-09-23T10:16:37Z (about 4 years ago)

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What is the difference between operator precedence and order of evaluation?

When doing something simple such as this

    int a=1;
    int b=2;
    int c=3;
    printf("%d\n", a + b * c);

then I was told that operator precedence guarantees that the code is equivalent to `a + (b * c)`, since `*` has higher precedence than `+`. And so the result is guaranteed to be 7 and not 9.

However, when I modify the above example like this:

    #include <stdio.h>

    int a (void) { printf("%s ",__func__); return 1; }
    int b (void) { printf("%s ",__func__); return 2; }
    int c (void) { printf("%s ",__func__); return 3; }

    int main (void)
    {
      printf("%d\n", a() + b() * c());
      return 0;
    }

Then I get the output `a b c 7`. How is this possible? And when I try the identical code on another compiler, I get the opposite order `c b a 7`.

Shouldn't operator precedence guarantee that `b()` is executed before `a()`?

c c++ operator-precedence order-of-evaluation unspecified-behavior

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