I wish to compare a SHA-256 hash which is stored in u8[32](after being calculated in kernel-space) with a 64 char string that the user passes as string.
For eg. : User passes a SHA-256 hash "49454bcda10e0dd4543cfa39da9615a19950570129420f956352a58780550839" as char* which will take 64 bytes. But this has to be compared with a hash inside the kernel space which is represented as u8 hash[32].
The hash inside the kernel gets properly printed in ASCII by the following code:
int i;
u8 hash[32];
for(i=0; i<32; i++)
printk(KERN_CONT "%hhx ", hash[i]);
Output : "49 45 4b cd a1 0e 0d d4 54 3c fa 39 da 96 15 a1 99 50 57 01 29 42 0f 95 63 52 a5 87 80 55 08 39 "
As the complete hash is stored in 32 bytes and printed as 64 chars in groups of 2 chars per u8 space, I assume that currently one u8 block stores information worth 2 chars i.e. 00101111 prints to be 2f.
Is there a way to store the 64 bytes string in 32 bytes so that it can be compared?
Here is how to use scanf to do the conversion:
char *shaStr = "49454bcda10e0dd4543cfa39da9615a19950570129420f956352a58780550839";
uint8_t sha[32];
for (int i = 0 ; i != 32 ; i++) {
sscanf(shaStr+2*i, "%2" SCNx8, &sha[i]);
printf("%02x ", sha[i]);
}
The approach here is to call sscanf repeatedly with the "%2" SCNx8 format specifier, which means "two hex characters converted to uint8_t". The position is determined by the index of the loop iteration, i.e. shaStr+2*i