Convert 4 bytes char to int32 in C

Question

I first convert an int32 number to char[4] array, then convert the array back to int32 by (int *), but the number isn't the same as before:

unsigned int num = 2130706432;
unsigned int x;
unsigned char a[4];

a[0] = (num>>24) & 0xFF;
a[1] = (num>>16) & 0xFF;
a[2] = (num>>8) & 0xFF;
a[3] = num & 0xFF;

x = *(int *)a;
printf("%d\n", x);

the output is 127. And if I set num = 127, the output is 2130706432. Does anyone have ideas?

Solution 1

Reverse the order of the a[] indexes, e.g,. a[0] -> a[3]

I think you have the endianness in reverse.

Try this:

a[3] = (num>>24) & 0xFF;
a[2] = (num>>16) & 0xFF;
a[1] = (num>>8) & 0xFF;
a[0] = num & 0xFF;

Solution 2

To see what happens use

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

to print both input and output number.

Endian-independent way:

x = (a[0] << 24) | (a[1] << 16) | (a[2] << 8) | a[3];

Solution 3

This line is never going to work correctly on a little-endian machine:

x = *(int *)a;

You need to unpack the data before you print out the value.

Solution 4

Your code a[0] = (num>>24) & 0xFF; takes the most significant 8 bits from num and sticks them in the first byte of a. On little endian machines the first byte holds the least signficant bits. That means that on little endian machines, this code takes the most significant 8 bits and stores them in the place where the least significant bits go, changing the value.

2130706432 is 0x7F000000 in hex, and 127 is 0x0000007F.

Also, x = *(int *)a; results in undefined behavior. Consider hardware where reading an int from an improperly aligned address causes a bus error. If a doesn't happen to be aligned properly for an int then the program would crash.

A correct approach to interpreting the bytes as an int would be std::memcpy(&x, a, sizeof x);

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Comments

[ninjalj]

Additionally, it may not work on big-endian machines with compilers that take advantage of strict-aliasing rules.

[rightaway717]

I don't know why this is voted up. As if it's an endian-independent way, it should work on my PC too. But I have to do quite the opposite uint32_t var2 = (v[3] << 24) | (v[2] << 16) | (v[1] << 8) | v[0];

[rightaway717]

I use vector<unsigned char> and fill it with uint32_t vars using memcpy: memcpy(&v[0], &var, sizeof var);

[Adam Trhon]

@rightaway717 So save it in endianess of your computer. To do it endian-independent way you have to convert uint32_t -> array as OP does and array -> uint32_t as I do.