Thread: Undefined behaviour doing an memcpy from long to float variable

I am not able to figure out the knowledge gap for the below piece of code.
I assume that the float variable to have an value equal to that of i , but it always prints 0.000. Am I missing something?

Code:

int main()
{


        unsigned long i =25,j=90;
        float d,m;
        short k;

        memcpy((void *)&d,(void *)&i,sizeof(float));
        memcpy((void *)&k,(void *)&i,sizeof(short));
        printf("Print......i:%d, d:%f, k:%d\n",i,d,k);

        return 0;
}

output
Print......i:25, d:0.000000, k:25


Thanks in Advace

Something to ponder...

Code:

#include <stdio.h>

int main(void)
{
    union fu {
        unsigned u;
        float f;
    } foo;

    foo.f = 25;

    printf("%x\n", foo.u);

    return 0;
}

Originally Posted by Hodor

Something to ponder...

Code:

#include <stdio.h>

int main(void)
{
    union fu {
        unsigned u;
        float f;
    } foo;

    foo.f = 25;

    printf("%x\n", foo.u);

    return 0;
}

Is it something to do with the way floating point numbers are represented?


But why typecasting it like this d=(float)i; gives an right value?

I am too confused......

Originally Posted by Salem

Yes, it's all about how floats are represented in memory.

Also, assignment makes the format conversion in a way that memcpy doesn't.

> (In Assignment scenario)Does that mean the compiler will take care of converting from float to int, which means that it take care of converting existing representation [S][Exponential][Mantissa] double/precision ?

> I came across this link
LLVM Project Blog: What Every C Programmer Should Know About Undefined Behavior #1/3

which recommends using memcpy instead of an assignment.
It is valid to use memcpy() for copying from float to unsigned long int.

> My understanding is since memcpy is byte wise copy now the unsigned long int variable contains an float represented data which is invalid, Is this true?


Thanks in advance

> (In Assignment scenario)Does that mean the compiler will take care of converting from float to int, which means that it take care of converting existing representation [S][Exponential][Mantissa] double/precision ?


> My understanding is since memcpy is byte wise copy now the unsigned long int variable contains an float represented data which is invalid, Is this true?