数据类型之间c的类型转换

Question

I wrote this small code to test my understanding. But didn't understand some facts behind it. I am working on a 64 bit little endian machine. So any pointer is 8 bytes. That means

#include<stdio.h>
int main(){
    char *c = (char *)0x12345678889;
    long a = 1;
    int b = (long)(c-a);
    /* int cc = (int)(c-a); gives compiler error */
    printf("val = %x and b = %x", c-a,b);
    return 0;
}
Output
val = 45678888 and b = 45678888

Say the starting address is 100. So the char* would be stored in memory as 100->89, 101->88 ... 105->12 and bytes 106 and 107 will be unused. Is this assumption of mine correct in the first place? Since int and long are 4 bytes in a 64 bit machine, it will start from 100,101,102 and 103 and will consider only these bytes. So 45678889 - 1 = 45678888. Is my understanding correct? Finally, I don't understand while the commented line gives a compiler error. The compiler had implicitly typecasted the above line. But why not below?

Answer 1

First you didn't know a-priori how values are stored, it depends on machine endianess . Values can be stored from lowest to highest digits or in the converse. Yours is from low to high.

Second, on many 64-bit machines int are 4 bytes long and long 8 bytes long. What happens is that your machine computes ca which is 0x12345678888 of type (char *) , then it convert it into a long with value 0x12345678888 and then truncate it silently (longs are silently converted to ints in C).

The comment gives a warning because ca is of type char * and that cannot be silently converted into an int (it is considered much dangerous dans long to int). Ok then you converted it explicitly but the compiler warns you that it is dangerous too. Note that these are warnings not errors (it may depend on compiler options...).