Typecasting from int,float,char,double

问题

I was trying out few examples on do's and dont's of typecasting. I could not understand why the following code snippets failed to output the correct result.

/* int to float */
#include<stdio.h>
int main(){
        int i = 37;
        float f = *(float*)&i;
        printf("\n %f \n",f);
        return 0;
}

This prints 0.000000

/* float to short */
#include<stdio.h>
int main(){
        float f = 7.0;
        short s = *(float*)&f;
        printf("\n s: %d \n",s);
        return 0;
}

This prints 7

/* From double to char */
#include<stdio.h>
int main(){
        double d = 3.14;
        char ch = *(char*)&d;
        printf("\n ch : %c \n",ch);
        return 0;
}

This prints garbage

/* From short to double */
#include<stdio.h>
int main(){
        short s = 45;
        double d = *(double*)&s;
        printf("\n d : %f \n",d);
        return 0;
}

This prints 0.000000

Why does the cast from float to int give the correct result and all the other conversions give wrong results when type is cast explicitly?

I couldn't clearly understand why this typecasting of (float*) is needed instead of float

int i = 10;
float f = (float) i; // gives the correct op as : 10.000

But,

int i = 10;
float f = *(float*)&i; // gives a 0.0000

What is the difference between the above two type casts?

Why cant we use:

float f = (float**)&i;
float f = *(float*)&i;

回答1:

In this example:

char ch = *(char*)&d;

You are not casting from double to a char. You are casting from a double* to a char*; that is, you are casting from a double pointer to a char pointer.

C will convert floating point types to integer types when casting the values, but since you are casting pointers to those values instead, there is no conversion done. You get garbage because floating point numbers are stored very differently from fixed point numbers.

回答2:

Read about the representation of floating point numbers in systems. Its not the way you're expecting it to be. Cast made through (float *) in your first snippet read the most significant first 16 bits. And if your system is little endian, there will be always zeros in most significant bits if the value containing in the int type variable is lesser than 2^16.

回答3:

If you need to convert int to float, the conversion is straight, because the promotion rules of C.
So, it is enough to write:

    int i = 37;
    float f = i;

This gives the result f == 37.0.

However, int the cast (float *)(&i), the result is an object of type "pointer to float".
In this case, the address of "pointer to integer" &i is the same as of the the "pointer to float" (float *)(&i). However, the object pointed by this last object is a float whose bits are the same as of the object i, which is an integer.

Now, the main point in this discussion is that the bit-representation of objects in memory is very different for integers and for floats.
A positive integer is represented in explicit form, as its binary mathematical expression dictates.
However, the floating point numbers have other representation, consisting of mantissa and exponent.

So, the bits of an object, when interpreted as an integer, have one meaning, but the same bits, interpreted as a float, have another very different meaning.

回答4:

The better question is, why does it EVER work. You see, when you do

typedef int T;//replace with whatever
typedef double J;//replace with whatever
T s = 45;
J d = *(J*)(&s);

You are basically telling the compiler (get the T* address of s, reintepret what it points to as J, and then get that value). No casting of the value (changing the bytes) actually happens. Sometimes, by luck, this is the same (low value floats will have an exponential of 0, so the integer interpretation may be the same) but often times, this'll be garbage, or worse, if the sizes are not the same (like casting to double from char) you can read unallocated data (heap corruption (sometimes)!).

来源：https://stackoverflow.com/questions/24270637/typecasting-from-int-float-char-double

标签

casting