convert uint8_t array to char array in c

Question

Initially I want to convert this uint8_t array to a char array in c. I have been a little stuck trying to resolve this problem. My first alternative solution is to copy another type value to the temporary one, copy the tmp value to a writable char, and then remove tmp value from memory. By the way this is used to accompany a blake hash function. Here is my code snippet:

char * bl(char *input)
{
    uint8_t  output[64];
    char msg[]= "";
    char *tmp;

    int dInt;

    memset(output,0,64);
    tmp = (char*) malloc(64);
    if (!tmp){
            exit( 1);
    }

    dInt = strlen(input);

    if (dInt > 0xffff){
            exit( 1);
    }
    uint8_t data[dInt];

    memset(data,0, dInt);
    strlcpy(data,input,dInt);
    uint64_t dLen =dInt;
    blake512_hash(output, data,dLen);

    int k;
    for (k=0;k<64;k++){
            tmp[k] = output[k];  //does this "copy" is buggy code?
    }

    memcpy(msg, tmp,64);
    //so here I can to delete tmp value
    // I dont want there were left unused value in memory
    // delete tmp;  
    free(tmp);

    return  msg;
}

I think the code above is still not efficient, so what are your opinion, hints and the fixes? Thank you very much before!

JosephH · Accepted Answer · 2012-04-18 02:18:40Z

7

First of all, you should never return a pointer to a local variable since the variable will be destroyed by the time the function exits. You should probably want to pass the output array to bl function and use that to output the string.

For most cases(if uint8_t IS char, which is usually the case), memcpy(msg, output, 64) should be sufficient. If you want to be strict about it(quite frankly blake512_hash shouldn't return uint8_t array in the first place if you are expecting char array as the output all the time), you could simply call msg[k] = (char)tmp[k] in your for loop and remove memcpy.

answered Apr 18, 2012 at 2:18

JosephH

8,8554 gold badges36 silver badges62 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

hafidh Over a year ago

I want to use this function later to be called in Python, that's way I'm expecting this funtion later will return an string value.

JosephH Over a year ago

@user1309539 Yes but you should still fix the problem i mentioned in my answer. You simply cannot return the pointer to a local variable.

hafidh Over a year ago

then tmp variable would be useless. what to be copied from tmp variable, since tmp variable is only allocated char not the output one ?

JosephH Over a year ago

@user1309539 you don't need tmp variable since you can simply copy the output to msg

Joop Eggen · Accepted Answer · 2012-04-18 02:23:36Z

0

A bit much is wrong here.

dInt = strlen(input) + 1; // dInt is the size of the string including the terminating '\0'.

strlcpy indeed uses the size, not strlen.

msg = tmp; and not freeing tmp. As msg is const char* "" (in C++ terms).

answered Apr 18, 2012 at 2:23

Joop Eggen

110k8 gold badges89 silver badges142 bronze badges

1 Comment

hafidh Over a year ago

umh, yes thanks. anymore wrong again, please give your suggestion about above code, thanks :)

Collectives™ on Stack Overflow

convert uint8_t array to char array in c

2 Answers 2

4 Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

4 Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related