C语言中scanf的变体

Question

scanf("%d %d"+2, &a, &b);
printf("%d\n%d", a, b);

It accepts only a and prints a and 0 .
Can anyone explain why this is happening? Also, if I write +1 instead of +2 , it accepts nothing and prints 0 and 0 .

Answer 1

This:

scanf("%d %d"+2,&a,&b);

is the same as

scanf(" %d", &a, &b);

which is the same as

scanf("%d", &a, &b);

which means that the extra &b argument is unnecessary.

What happens here is that "%d %d" is a char* . Adding two to it results into a pointer pointing two bytes ahead which means that it now points to " %d" . The leading space is unnecessary because %d already skips leading whitespace characters.

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When you use +1 instead of +2 , the scanf is the same as

scanf("d %d", &a, &b);

which means that it expects a d in the input followed by an integer to be assigned to a . Since you provide a number instead of d in the input, the scanf fails and returns 0. Thus, nothing is accepted and the execution reaches the printf which prints the value of both a and b .