I need to create a malloc array of strings and print those strings out

Question

important: I don't have a deep understanding of malloc so try to put things as simple as possible Greetings, I want to create a dynamic array of strings (using malloc) and then print those strings out (for starters).

That's what I have

int main(){
    char **namesofcolumns = malloc(4 * sizeof(char*));
    for(int i = 0; i < 4; i++){
        namesofcolumns[i] = malloc(20 * sizeof(char));
    }
    int count = 1;
    for(int i = 0; i < 4; i++){
        printf("Enter %d column name\n", count);
        scanf("%s", namesofcolumns[i]);
        count++;
    }

I probably have errors here already. How do I print entered strings out?

Printf wants a char*. Ask yourself what the types are of namesofcolumns and namesofcolumns[i] — Botje
– Botje, Commented Mar 31, 2024 at 12:32
@xing, thanks a lot. Now I need to return this array namesofcolumns and than use it in a function that will print out a table header ''' void printHeader(char namesOfRows[4][20]){ printf("__________________________________\n"); for(int itemNo = 0; itemNo < 4; itemNo++){ for(int itemLen = 0; itemLen < strlen(namesOfRows[itemNo]); itemLen++){ printf("%c", namesOfRows[itemNo][itemLen]); } printf(" "); printf("|"); printf(" "); } } ''' how do I return this array namesofcolumns? — Dark Dev
– Dark Dev, Commented Apr 2, 2024 at 9:44

allCodeBreaker · Accepted Answer · 2024-07-12 08:22:11Z

0

namesofcolumns have 4 string pointers so you have to derefence it access the value stored in namesofcolumns.Add the following code to you program to see expected result.

count = 1;
for(i = 0;i < 4;i++){
    printf("Name %d :%s\n",count,*namesofcolumns);
    namesofcolumns++;
    count++ ;
}

answered Jul 12, 2024 at 8:22

allCodeBreaker

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1 Comment

Lundin Over a year ago

Since none but namesofcolumns point at the allocated memory, this will create memory leaks.

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