Forgotten code for user input in C in Linux Terminal - c

Question

I've written the code, however I cannot remember how to get the user's input.
The command to run the code is ./rle "HHEELLLLO W" but in my code, I don't know how to get the code to read the "HHEELLLLO W".
If I have a regular printf function, the code will work and will print H1E1L3O 0W0 but I want it so any user's input can be calculated.
I tried using atoi but I think I've done it wrong and I don't know how to fix it.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX_RLEN 50
char* encode(char* src)
{
int rLen;
char count[MAX_RLEN];
int len = strlen(src);
char* dest = (char*)malloc(sizeof(char) * (len * 2 + 1));
int i, j = 0, k;
for (i = 0; i < len; i++) {
dest[j++] = src[i];
rLen = 1;
while (i + 1 < len && src[i] == src[i + 1]) {
rLen++;
i++;
}
sprintf(count, "%d", rLen);
for (k = 0; *(count + k); k++, j++) {
dest[j] = count[k];
}
}
dest[j] = '\0';
return dest;
}
int main(int argc, char ** argv)
{
char str[] = atoi(argv[1]);
char* res = encode(str);
printf("%s", res);
getchar();
}
When I compile it, I get this error:
rle.c: In function ‘main’:
rle.c:47:18: error: invalid initializer
char str[] = atoi(argv[1]);
^~~~
rle.c:45:14: error: unused parameter ‘argc’ [-Werror=unused-parameter]
int main(int argc, char ** argv)
^~~~
cc1: all warnings being treated as errors

Answer 1

atoi converts strings (of digits) to integers, which is nothing like what you to do here.
I would recommend doing this one of two ways:
(1) Use the command-line argument directly:
int main(int argc, char ** argv)
{
char* str = argv[1];
char* res = encode(str);
printf("%s\n", res);
}
This works because argv[1] is already a string, as you want. (In fact in this case you don't even need str; you could just do char* res = encode(argv[1]);.)
In this case you will have the issue that the shell will break the command line up into words as argv[1], argv[2], etc., so argv[1] will contain just the first word. You can either use quotes on the command line to force everything into argv[1], or use the next technique.
(2) Read a line from the user:
int main(int argc, char ** argv)
{
char str[100];
printf("type a string:\n");
fgets(str, sizeof(str), stdin);
char* res = encode(str);
printf("%s\n", res);
}
In both cases there's also some additional error checking you theoretically ought to do.
In the first case you're assuming the user actually gave you a command-line argument. If the user runs your program without providing an argument, argv[1] will be a null pointer, and your code will probably crash. To prevent that, you could add the test
if(argc <= 1) {
printf("You didn't type anything!\n");
exit(1);
}
At the same time you could double-check that there aren't any extra arguments:
if(argc > 2) {
printf("(warning: extra argument(s) ignored)\n");
}
In the second case, where you prompt the user, there's still the chance that they won't type anything, so you should check the return value from fgets:
if(fgets(str, sizeof(str), stdin) == NULL) {
printf("You didn't type anything!\n");
exit(1);
}
As you'll notice if you try this, there's also the issue that fgets leaves the \n in the buffer, which may not be what you want.