how to print contents of a void *?

Question

static inline uint256_t sha256(const void *data, size_t len)
{
    secp256k1_sha256_t cxt;
    secp256k1_sha256_initialize(&cxt);
    secp256k1_sha256_write(&cxt, (uint8_t *)data, (int)len);
    uint256_t res;
    secp256k1_sha256_finalize(&cxt, (uint8_t *)&res);
    return res;
}

I want to see what's inside data and what's being returned by the above function, so printing seems like an easy way but how do i print contents of *data not the pointer to data because *%p* gave me an address.

score 2 · Accepted Answer · answered Jan 15 '23 at 07:06

2

Use a loop.

const unsigned char *bytes = data;
for (size_t i = 0; i < len; i++) {
  printf("%02X ", bytes[i]);
}

answered Jan 15 '23 at 07:06

chux - Reinstate Monica

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Allan Wind · Answer 2 · 2023-01-15T07:03:47.673

0

If you want to print the pointer address:

printf("%p", data);

If you want to print the value (data), then you need to cast the pointer to something else, probably a (const char *). Used a "%s" which would work well if data is string:

printf("%*s", (int) len, (const char *) data);

If your data is binary or contain '\0' bytes then you probably hex encoding instead.

edited Jan 15 '23 at 07:03

answered Jan 15 '23 at 06:57

Allan Wind

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`"%*s"` work well with a _string_, as it stops printing when a byte 0 encountered - or `len` bytes. To certainly print `len` bytes takes other code. – chux - Reinstate Monica Jan 15 '23 at 07:02
@chux-ReinstateMonica Yup, I will spell it out even more. – Allan Wind Jan 15 '23 at 07:03

how to print contents of a void *?

2 Answers2