(unsigned long)1 和 *(unsigned long *) 是什么意思？

Question

This is a program that prints floating point integers bit representation. What does *(unsigned long *)a and (unsigned long)1 mean?

#include <stdio.h>

void printBits(void *a){

    int i;
    unsigned long x;
    x = *(unsigned long *)a;
    for (i = 63; i >= 0; i--){
        if ((x & ((unsigned long)1 << i)) != 0)
            printf("1");
        else
            printf("0");
        if (i == 63)
            printf(" ");
        if (i == 62)
            printf(" ");
        if (i == 52)
            printf(" ");
    }
    printf("\n");
}

int main(void){
    
    double x = -145.4;
    printBits(&x);
    return 0;
}

Answer 1

Both are type casts ie explicit type conversions.

---

In

x = *(unsigned long *)a; 
    ^                
//dereference

(unsigned long *)a casts a to a pointer to unsigned long and then dereferences it to assign its value to the local variable x .

---

(unsigned long)1

Simply casts 1 to unsigned long which by default would have the type int .

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The first is needed because you can't dereference a void pointer, the second is there to avoid undefined behavior, see chux - Reinstate Monica's comment . More details here .