为什么分配时X的值不增加

Question

I'm new to C language and I'm not entirely sure why on Line 4 when post-increment is done the value of x isn't change what I mean is that

x = printf("%d",x++);

The value for x was 12 so the printf would print 12 and then x should be assigned 2 and while ++ was there x should be later changed with 2+1 and on line 6 pre-increment is done so output shouldn't be 124 .

Why x on line 4 isn't added?

Please help.

#include <stdio.h>

int main(){

int x = 12;

x = printf("%d", x++);

printf("%d", ++x);

return 0;

}

Answer 1

Make yourself aware of sequence point . From this [emphasis mine] :

There is a sequence point after the evaluation of all function arguments and of the function designator, and before the actual function call .

From this [emphasis mine] :

Increment operators initiate the side-effect of adding the value 1 of appropriate type to the operand. Decrement operators initiate the side-effect of subtracting the value 1 of appropriate type from the operand. As with any other side-effects, these operations complete at or before the next sequence point .

Looks at this statement:

x = printf("%d", x++);

The post increment operator increase the value of operand by 1 but the value of the expression is the operand's original value prior to the increment operation .
So, the value of x passed to printf() will be its original value which is 12 and due to sequence point, before calling printf() the value of x will be incremented by 1 . The return value of printf() will be assigned to x which overwrites the last value of x which is the incremented value due to post ++ operator. Hence, after this statement the value of x is 2 .

Answer 2

The value for x was 12 so the printf would print 12 and then x should be assigned 2 and while ++ was there x should be later changed with 2+1 and on line 6 pre-increment is done so output shouldn't be 124.

no, the assignment is done after all concerning printf("%d", ++x); , your code is equivalent to that :

#include <stdio.h>

int main(){
  int x = 12;

  int y = printf("%d", x++);

  x = y;

  printf("%d", ++x);

  return 0;
}

so x = printf("%d", ++x); does :

• printf writes 12
• then x is incremented to value 13
• then x is assigned to the result of printf valuing 2

then you execute printf("%d", ++x); while x values 2 before, so x is incremented before to be given in argument, so 3 is written

and the final print result is 123

PS. As said by @HS in an other remark :There is a sequence point after the evaluation of all function arguments (x++ is an argument to printf()) and the pre/post increment/decrement operation complete at or before the next sequence point.

Answer 3

This will work for you:

++x is prefix increment and the x++ is postfix increment, and here is the difference on both of them:

The prefix increment returns the value of a variable after it has been incremented.

On the other hand, the more commonly used postfix increment returns the value of a variable before it has been incremented.

#include <stdio.h>

int main()
{
    int x = 12;

    x = printf("%d", ++x);

    printf("%d", ++x);

    return 0;
}

Answer 4

x= printf("%d", x++); This is about the size of that integer. x++ is in printf it is printing the value of x and increase the value by one. But it doesn't affect the size of that integer. Then you assign the size of x to x not the value. That's why that incremental had not worked there.

I think you can understand what I am saying.