Why double b=nextafter(0., 1.) is 0.000000e+00 for float type?

I found the result of nextafter(0., 1.) with data type double and float are not the same.

#include <math.h>
#include <stdio.h>

int main(){
    double a;
    float b;
    a=nextafter(0., 1.);
    b=nextafter(0., 1.);
    printf("default %e\n",nextafter(0., 1.));
    printf("double %e\n",a); //double 4.940656e-324
    printf("float %e\n",b); //float 0.000000e+00
    return 0;
}

Why the smallest float bigger than 0.0 is 0.000000e+00 ?

When you call nextafter(0., 1.) and store the result in a float, implicit type conversion happens. The function returns the smallest positive subnormal double, which is 4.940656 x 10⁻³²⁴. This value is then rounded to float, but it's too small to be represented as a float, leading to zero.

To get the correct float result, you should use nextafterf(0.f, 1.f):

float b = nextafterf(0.f, 1.f);

This will give you the smallest positive subnormal float.