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Comments on What is the difference between operator precedence and order of evaluation?
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What is the difference between operator precedence and order of evaluation?
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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()?
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