CODEWITHSUNDEEP
cc-strings
user3603183
user3603183

Reputation: 333

Appending an int array to a string in c:

I'm trying to append an array of integers into a string, basically I'm trying to get it into the following format: A :123456 where A is a character, given by a parameter, and 123456 is an integer array. I'm getting a bus error when running code, and I've done a search on here, and on google for the correct way to do something similar to what I have been hoping to achieve, but the examples don't directly relate to this so I was wondering if I could get some help. Thanks in advance

char *print_cards_played(Player *player) {
    char *result = "";
    int i = 0;

    result += sprintf(result, "%c :", player->playerID);

    while(player->cardsPlayed[i] != 0) {
        result += sprintf(result, "%d", player->cardsPlayed[i]);
        i++;
    }

    result += sprintf(result, "\n");

    return result;
}

Upvotes: 1

Views: 1038

Answers (5)

unwind
unwind

Reputation: 399793

You cannot "append" to a string defined by:

char *result = "";

All that gives you is a pointer named result which points at a single read-only location in memory where the character '\0' (the string terminator) is stored. You can't write to it.

You need to have an actual buffer of some suitable size:

char *result = malloc(128);

If that allocation succceeds, you have what is called a "heap buffer" which will stay around even if the function that did the allocation exits. You can safely return this pointer, unlike a local array which would disappear as the function exited.

Then you can use sprintf(), but of course not with +=, you can't add strings in C. The best solution might be to sprintf() to a temporary small string, then strcat() that string onto the result. You can keep track of the appended length to make it fast, but that's not critical for small problems on a modern machine.

Upvotes: 3

M Oehm
M Oehm

Reputation: 29126

You could write your own function and model it on the snprintf paradigm, which takes a buffer and its length and fills that buffer with the desired string. If the buffer is too small, the string will be truncated, but still be a valid, null-terminated string:

int str_cards_played(char *buf, int nbuf, const Player *player)
{
    int n;
    int i = 0;

    n = snprintf(buf, nbuf, "%c: ", player->playerID);

    while(n < nbuf && player->cardsPlayed[i] != 0) {
        n += snprintf(buf + n, nbuf - n, "%d", player->cardsPlayed[i]);
        i++;
    }

    if (n < nbuf) n += snprintf(buf + n, nbuf - n, "\n");
    return n;
}

You'd call the function like so:

char buf[20];

str_cards_played(buf, sizeof(buf), player);
printf("'%s'\n", buf);

If the allocation is on the stack like here, you don't have to worry about memory management.

Upvotes: 1

BLUEPIXY
BLUEPIXY

Reputation: 40145

try this

char *print_cards_played(Player *player) {
    char result[128];// = "";
    char *p = result;
    int i = 0;

    p += sprintf(p, "%c :", player->playerID);

    while(player->cardsPlayed[i] != 0) {
        p += sprintf(p, "%d", player->cardsPlayed[i]);
        i++;
    }

    p += sprintf(p, "\n");

    return strdup(result);
}

Upvotes: 2

l0gg3r
l0gg3r

Reputation: 8954

Use strcatInt described below, to concat char* and int

char* itoa(int i, char b[]){
    char const digit[] = "0123456789";
    char* p = b;
    if(i < 0){
        *p++ = '-';
        i *= -1;
    }
    int shifter = i;
    do{ //Move to where representation ends
        ++p;
        shifter = shifter/10;
    } while(shifter);
    *p = '\0';
    do{ //Move back, inserting digits as u go
        *--p = digit[i%10];
        i = i/10;
    } while(i);
    return b;
}

char* strCatInt(char* str, int num) {
    char numStr[10];
    itoa(num, numStr);
    return strcat(str, numStr);
}

Upvotes: 0

Sunil Bojanapally
Sunil Bojanapally

Reputation: 12658

Bus error is because of: char *p = "String Literal";

The reason stated clearly in c-faq as,

Anywhere else, it turns into an unnamed, static array of characters, and this unnamed array may be stored in read-only memory, and which therefore cannot necessarily be modified. In an expression context, the array is converted at once to a pointer, as usual, so the second declaration initializes p to point to the unnamed array's first element.

Upvotes: 0

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