C 将 char* 转换为 uint8_t []

Question

I'm working on a program that receives an encrypted string. For this I'm using tiny-AES-c library.

The encrypted string I'm receiving is char* . In order to use the string with the function AES_CBC_decrypt_buffer(struct AES_ctx* ctx, uint8_t* buf, uint32_t length) I will need to convert the string to uint8_t [] .

This is how the tiny-AES-c library likes the format, as seen in their test.c :

uint8_t in[]  = { 0x6b, 0xc1, 0xbe, 0xe2, 0x2e, 0x40, 0x9f, 0x96, 0xe9, 0x3d, 0x7e, 0x11, 0x73, 0x93, 0x17, 0x2a };

I've been told this can be achieved using strtoul . I've researched this function quite a bit but can't seem to find any questions/documentation/examples related to the issue I'm having.

My question being, how can I convert the contents of a char* to an uint8_t [] array, in the format like the snippet above?

Answer 1

This could be as simple as:

char* buf = "test";
size_t length = 5;

AES_CBC_decrypt_buffer(ctx, (uint8_t*) buf,  (uint32_t) length)

uint8_t and char are largely the same thing for binary data, same as size_t and uint32_t are compatible for values under 2^32.

C99 code that uses the integer types will use things like uint8_t , but older C code, or code with no C99 dependency, will use char . They basically mean the same thing.