size of uint8, uint16 and uint32?

Question

I am working with code that uses types like uint8 (1 byte wide unsigned integer), uint16 (2 byte wide unsigned integer), uint32 (4 byte wide unsigned integer), etc..

My questions are:

• Are uint8 and char the same size?

• Are uint16 and int the same size?

• Are uint32 and long the same size?

• Are uint64 and double the same size?

for example:

uint8 c[20];

Shouldn't the size of this be 20?

I wrote some code in visual studio:

#include <stdio.h>
#include <string.h>
typedef unsigned int uint32;
typedef unsigned int uint8;

int main()
{
    double a = 1320.134;
    uint32 b;
    uint8 c[20];
    b = (unsigned int)a;
    c[3] = b;
    printf("value is %d", c[3]);
    return 1;
}

but the size of c in debug mode is 50. Why is that?

Answer 1

uint8, uint16, uint32, and uint64 are probably Microsoft-specific types.

As of the 1999 standard, C supports standard typedefs with similar meanings, defined in <stdint.h>: uint8_t, uint16_t, uint32_t, and uint64_t. I'll assume that the Microsoft-specific types are defined similarly. Microsoft does support <stdint.h>, at least as of Visual Studio 2010, but older code may use uint8 et al.

The predefined types char, short, int et al have sizes that vary from one C implementation to another. The C standard has certain minimum requirements (char is at least 8 bits, short and int are at least 16, long is at least 32, and each type in that list is at least as wide as the previous type), but permits some flexibility. For example, I've seen systems where int is 16, 32, or 64 bits.

char is almost always exactly 8 bits, but it's permitted to be wider. And plain char may be either signed or unsigned.

uint8_t is required to be an unsigned integer type that's exactly 8 bits wide. It's likely to be a typedef for unsigned char, though it might be a typedef for plain char if plain char happens to be unsigned. If there is no predefined 8-bit unsigned type, then uint8_t will not be defined at all.

Similarly, each uintN_t type is an unsigned type that's exactly N bits wide.

In addition, <stdint.h> defines corresponding signed intN_t types, as well as int_fastN_t and int_leastN_t types that are at least the specified width.

The [u]intN_t types are guaranteed to have no padding bits, so the size of each is exactly N bits. The signed intN_t types are required to use a 2's-complement representation.

Although uint32_t might be the same as unsigned int, for example, you shouldn't assume that. Use unsigned int when you need an unsigned integer type that's at least 16 bits wide, and that's the "natural" size for the current system. Use uint32_t when you need an unsigned integer type that's exactly 32 bits wide.

(And no, uint64 or uint64_t is not the same as double; double is a floating-point type.)

Answer 2

It's quite unclear how you are computing the size ("the size in debug mode"?").

Use printf():

printf("the size of c is %u\n", (unsigned int) sizeof c);

Normally you'd print a size_t value (which is the type sizeof returns) with %zu, but if you're using a pre-C99 compiler like Visual Studio that won't work.

You need to find the typedef statements in your code that define the custom names like uint8 and so on; those are not standard so nobody here can know how they're defined in your code.

New C code should use <stdint.h> which gives you uint8_t and so on.