Printf sequence point

Question

the output of the next code is:

40
30
20
10

 unsigned char numbers[] = {10,20,30,40};
 unsigned char* ptr = numbers;
 printf("%d\n%d\n%d\n%d",*ptr, *(ptr++), *(ptr++), *(ptr++) );

I think ++ operators are done first, so I understand why the first value printed is 40, but how it comes to print 30, 20 and 10 after it ? It's going backwards this way!

When you say "++ operators are done first", what do you mean? They're done before what? — David Schwartz, Dec 08 '18 at 00:12

score 0 · Answer 1 · answered Dec 07 '18 at 23:45

0

What you invoke is undefined behavior.

In C there is no sequence point between the evaluation of function parameters.

You have to write:

printf("%d\n%d\n%d\n%d",*ptr, *(ptr), *(ptr+1), *(ptr+2) );
ptr += 3;

answered Dec 07 '18 at 23:45

Osiris

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Printf sequence point

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