为什么我的程序的output 0.000000 应该是我在提示时输入的值。 下面是代码

Question

I am trying to execute this code. but the number variable isn't storing the value I input,what could be posssible explanations of this behaviour.

#include <stdio.h>
int main(){
    float number;
    
    printf("enter a value\n");
    scanf("%5.3f",&number);
    printf("u entered : %f",number);

    return 0;
}

and below is sample of its execution

enter a value
78.467  
u entered : 0.000000

Why isn't number storing the value 78.467

Answer 1

• %5.3f is not a valid scanf format. It should be %f .
• You also do not check that scanf succeeds. Always do.

Example:

#include <stdio.h>
int main() {
    float number;

    puts("enter a value");
    if(scanf("%f", &number) == 1) {          // check that it succeeds
        printf("u entered : %5.3f", number); // in printf %5.3f is ok
    } else {
        puts("Erroneous input");
    }
}

Answer 2

"%5.3f" in scanf("%5.3f",&number); is an invalid format so the result is undefined behavior (UB).

The "5" is OK. It limits the number of characters read in the conversion to at most 5.

".3" is not part of a valid conversion specification.

Save time, enable all warnings. @David Ranieri

// scanf("%5.3f",&number);
if (scanf("%f",&number) != 1) {
  fprintf(stderr, "Non-numeric input\n");
  return -1; 
}