Using array of string in C

Question

I am just practicing to understand dynamic allocation in C. I am getting a segmentation fault error. I am not sure where I made mistake.

int wordcount = 5;
char **args = (char**)malloc(wordcount * sizeof(char*));

for ( int i = 0; i < wordcount; i++) {
    args[i] = (char*)malloc(167 * sizeof(char));
}

int c=0;
while(c < wordcount){
    strcpy("hello\n",  args[c]);
    c++;
}

"not sure where I made mistake" a debugger would point you to the line crashing the program. — alk
– alk, Commented May 19, 2018 at 16:56
It seems all you need is a good strcpy reference. By the way, just about any tutorial or book would have told you how to use strcpy already. — Some programmer dude
– Some programmer dude, Commented May 19, 2018 at 16:56
strcpy("hello\n", args[c]); ==> strcpy(args[c], "hello\n"); — 4386427
– 4386427, Commented May 19, 2018 at 16:59
Well no, char ** args; defines args to be a pointer to pointer to char. Also there is no data type "string" in C. — alk
– alk, Commented May 19, 2018 at 17:09

Josu Goñi · Accepted Answer · 2018-05-19 18:18:35Z

1

You are trying to copy args[c] to the location of the "hello\n" string, which is in read only memory, you should change the order of the parameters.

answered May 19, 2018 at 18:18

Josu Goñi

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