谁能解释以下程序的扫描值输出？

Question

#include<stdio.h>

int main()
{
    int i;
    printf("%d\n", scanf("%d", &i) );
    return 0;
}

If some value is supplied to scanf() the output of the program is still 1. Why ?

Answer 1

if some value is supplied scanf but the output of the program is still 1. Why?

Because scanf returns an int as per the documentation . scanf will return the total number of items scanned and assigned successfully.

In your case, scanf will return 1 if it was successful in scanning an int from the stdin , else it will return 0. It will also return -1 on encountering EOF . Then, this value is printed by the printf you have.

---

If you want to print the value of i , seperate the scanf and the printf , ie, use

int i;
scanf("%d", &i)
printf("%d\n", i);

instead of

int i;
printf("%d\n" , scanf("%d", &i) );

Answer 2

In the below statement,

printf("%d\n" , scanf("%d", &i) );

you're not printing the value scanned and stored by scanf() , you're printing the return value of scanf() .

The value, scanned by scanf() will be stored in the supplied argument, here, the variable i .

Remember, scanf() does not return the scanned value , it returns the number of items it successfully matched and assigned .

If you want to print the scanned value, you have to use the same variable where the value is stored, ie, i .

  printf("The scanned value is %d\n", i);

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That said, as a note, the recommended signature for main() is int main(void) .

Answer 3

The return type of scanf() is number of items of the argument list successfully filled.In your case one item is being read so if successful 1 will be returned by scanf() Similarly the following would return 2 because there are two items being read.

 int i,j;
 printf("%d\n" , scanf("%d %d", &i,&j) );