how to iterate through argv through command line and C - c

Question

#include <stdio.h>
int main(int argc, char **argv) {
if (argc != 2) {
fprintf(stderr, "usage: one file only", argv[0]);
return (1);
} else {
for(int i = 0; i < argc; i++) {
putchar(argv[i]);
}
}
}
Say I want to print the input it takes in, for instance
$ gcc -Wall fileabove.c
$ ./a.out abcdefghijlmn
abcdefghijlmn
Basically just prints out whatever text I put into it.

Answer 1

putchar(argv[i]);
is incorrect, because putchar expects a single character (type char), you are passing a pointer (char*). It should be
puts(argv[i]);
So the correct code:
#include <stdio.h>
int main(int argc, char **argv) {
if (argc != 2) {
fprintf(stderr, "usage: one file only\n", argv[0]);
return 1;
}
// no need for the else
// you exit program anyway if argc != 2
// makes code more readable
puts(argv[1]);
return 0;
}
If you want to print character by character:
#include <stdio.h>
int main(int argc, char **argv) {
if (argc != 2) {
fprintf(stderr, "usage: one file only\n", argv[0]);
return (1);
}
for(size_t i = 0; argv[1][i] != '\0'; ++i)
putchar(argv[1][i]);
putchar('\n');
return 0;
}
edit changed puts("") to putchar('\n') and removed strlen as Jonathan Leffler mentioned in the comments.

Answer 2

The argument array is always in the format of first index being the called program name and the subsequent index values being the parameters in order. Because you only want one argument to process, I have simplified your code as follows:
#include <stdio.h>
int main(int argc, char **argv) {
if (argc != 2) {
fprintf(stderr, "%s usage: one file only\n", argv[0]);
return 1;
} else {
printf("%s\n",argv[1]); //print only parameter and new line
}
return 0;
}
When the user doesn't specify one parameter, then the error pops up including the program name. When the parameter is specified, it is stored in the second index of the array (index 1) and in my code, I simply printed it out using printf.