简单地在C中使用printf和doublef的scanf的问题

Question

I've made extremely easy program:

#include <stdio.h>
int main()
{
    double x;
    printf("Write your number \n");
    scanf ("%f", &x);
    printf("You've written %f \n", x);

    return 0;
}

And as a result strange number appears (no matter what x I give): "You've written 83096261053132580000000000000000000000000000000000000000000 "

What's wrong with this? This program works fine when I change all numbers into an 'int' type.

Answer 1

See what happens when you compile it with warnings enabled:

amb@nimrod-ubuntu:~/so$ gcc -Wall x.c -o x
x.c: In function ‘main’:
x.c:6:9: warning: format ‘%f’ expects argument of type ‘float *’, but argument 2 has type ‘double *’ [-Wformat]

Change %f to %lf and the scanf and printf functions will correctly take a double .

Answer 2

Wrong specifier for scanf() .

With scanf() , "%f" matches a float * , yet coded passed a double * . Use "%lf" instead. Code's printf() is fine. See Correct format specifier for double in printf

 double x;
 printf("Write your number \n");
 // scanf ("%f", &x);
 scanf ("%lf", &x);
 printf("You've written %f \n", x);

A good compiler should have warned of your incorrect code as suggested by @abligh. Either enable all warnings or consider a new compiler.

---

When scanning, the &x must match the proper scanf() print specifier.

• "%f" matches a float *
• "%lf" matches a double *

The using printf() things are easier. If a float or double is passed, being a variadic function, float is promoted to double .

• "%f" and "%lf" match a double

Answer 3

%f is used for float type. You should use %lf for double (Long Float).

Answer 4

You need to use %lf for scanf and printf.

In other words, this:

#include <stdio.h>
int main()
{
    double x;
    printf("Write your number \n");
    scanf ("%lf", &x);
    printf("You've written %lf \n", x);

    return 0;
}