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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)...
#4: Post edited
by
Lundin
·
2021-02-01T11:54:34Z (almost 4 years ago)
Formatting
- 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 (over 4 years ago)
#2: Post edited
by
Lundin
·
2020-09-23T14:17:28Z (over 4 years ago)
- 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 (over 4 years ago)
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()`?