unexpected result from printf("%d",*p++);

Asked Aug 23 '21 at 00:26

Active Aug 23 '21 at 00:26

Viewed 47 times

#include <stdio.h>

int main(){
   int a[3] = {10,20,30};
   int *p;
   p = a;
   printf("%d %d %d\n", a[0], a[1], a[2]);
   printf("%d %d %d\n", p[0], p[1], p[2]);
   printf("%d %d %d\n", *p, *(p+1), *(p+2));
   printf("%d %d %d\n", *p++, *p++, *p++);
}

expected result
10 20 30
10 20 30
10 20 30
10 20 30

result
10 20 30
10 20 30
10 20 30
30 20 10

asked Aug 23 '21 at 00:26

박고은

Probably because the parser and generator processes parameters right to left. – LoztInSpace Aug 23 '21 at 00:33
@dbush the dupe is incorrect. Sequence points are defined here, order of evaluation of parameters is not – 0___________ Aug 23 '21 at 00:40
1

@0___________: What sequence points do you think there are in `printf("%d %d %d\n", *p++, *p++, *p++);`? – Eric Postpischil Aug 23 '21 at 00:41
1

@0___________ The dup specifically addresses this case. – dbush Aug 23 '21 at 00:42
@박고은 No one can say how `printf("%d %d %d\n", *p++, *p++, *p++)` is evaluated. You may have thought it would be evaluated left-to-right, but no: it might be evaluated in any order, and in fact it's undefined. – Steve Summit Aug 23 '21 at 01:20

unexpected result from printf("%d",*p++);

0 Answers0