why the result of 's-t" in the while loop is "67"?

Question

please i can't understand the while loop , why the loop show A two time and not one

char *s,*t;
s="ABC";
t="AA";
do {printf("A");}
while(*s++==*t++);
printf("%d %d",*s-*t);

So which one is the question? The displaying A twice or the 67? — Sami Kuhmonen, Sep 15 '15 at 17:09
Indeed - one question per post, please. For the "why does it display twice" you should think consider why you're using `do/while` instead of `while`... and what the result of `*s++` is. — Jon Skeet, Sep 15 '15 at 17:10

score 1 · Answer 1 · answered Sep 15 '15 at 17:08

1

In your code

printf("%d %d",*s-*t);

invokes undefined behavior as you're not supplying enough argument to sate the supplied format specifiers.

answered Sep 15 '15 at 17:08

Sourav Ghosh

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score 1 · Answer 2 · answered Sep 15 '15 at 17:15

why the loop show A two time and not one?

The first time after the do, the print occurs unconditionally. Then the test occurs, and the condition is true (A == A). So the loop starts again, printing A the second time. After that, the test occurs again, and the condition is false (D != A), so the loop terminates.

score 1 · Answer 3 · answered Sep 15 '15 at 17:22

1

It prints 67 because (a) you are using the %d decimal number output format, and (b), after the loop ends, *s == 'C' and *t == '\0', and the difference of their ASCII values is 67.

answered Sep 15 '15 at 17:22

Weather Vane

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score 0 · Accepted Answer · answered Sep 15 '15 at 17:25

A do...while loop always runs at least once. So on the first iteration, A gets printed. In the condition s points to the first character in its string (A), and t points to the first character in its string ( also A). So it compares ('A'=='A') which is true, then s and t are both incremented to point to the second character in each string (A and B respectively).

On the second iteration, another A gets printed. In the condition it compares ('B'=='A') which is false, then s and t are both incremented to point to the third character in each string. Since the t string only contains two characters, t actually points to the NULL byte at the end of the string.

In the printf, it subtracts the value pointed to by s (C) by the value pointed to by t (a NULL byte, which in your implementation is 0). The ASCII value of C is 67, and 67 - 0 = 0, so 67 is printed.

why the result of '*s-*t" in the while loop is "67"?

4 Answers4

why the result of 's-t" in the while loop is "67"?