How to use atoi with an int and malloc? [closed] - c

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When I try and use atoi with an int and malloc I get a bunch of errors and key is given the wrong value, what am I doing wrong?
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct arguments {
int key;
};
void argument_handler(int argc, char **argv, struct arguments *settings);
int main(int argc, char **argv) {
argv[1] = 101; //makes testing faster
struct arguments *settings = (struct arguments*)malloc(sizeof(struct arguments));
argument_handler(argc, argv, settings);
free(settings);
return 0;
}
void argument_handler(int argc, char **argv, struct arguments *settings) {
int *key = malloc(sizeof(argv[1]));
*key = argv[1];
settings->key = atoi(key);
printf("%d\n", settings->key);
free(key);
}

Answer 1

You probably want this:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct arguments {
int key;
};
void argument_handler(int argc, char** argv, struct arguments* settings);
int main(int argc, char** argv) {
argv[1] = "101"; // 101 is a string, therefore you need ""
struct arguments* settings = (struct arguments*)malloc(sizeof(struct arguments));
argument_handler(argc, argv, settings);
free(settings);
return 0;
}
void argument_handler(int argc, char** argv, struct arguments* settings) {
char* key = malloc(strlen(argv[1]) + 1); // you want the length of the string here,
// and you want char* here, not int*
strcpy(key, argv[1]); // string needs to be copied
settings->key = atoi(key);
printf("%d\n", settings->key);
free(key);
}
But this is very awkward, actually the argument_handler can be rewritten like this:
void argument_handler(int argc, char** argv, struct arguments* settings) {
settings->key = atoi(argv[1]);
printf("%d\n", settings->key);
}
Disclaimer: I only corrected what was obviously wrong, there are still checks that need to be done, e.g. if argc is smaller than 2 etc.