Argument processing

Learn to process arguments by looping over argv.

Loop over arguments

To process arguments you set up a for loop from argv[1] to argv[argc-1]. At each step argv[i] gives the corresponding string.

Example: for (int i = 1; i < argc; i++) printf("%s\n", argv[i]);. The loop starts at i = 1 because argv[0] is the program name and is usually skipped. The %s conversion prints a string. This pattern underlies tools that take multiple inputs such as file lists or search terms.

Simulated arguments

To try the argument-processing pattern without run-time input, declare a local array resembling argv. This keeps the output always the same.

Example: char *args[] = {"prog", "apple", "pear"}; int n = 3;. Then for (int i = 1; i < n; i++) printf("%s\n", args[i]);. This technique helps when testing command-line tools: put a fixed array in place of real argv and verify the expected output. The same loop pattern works for both simulation and real arguments.

Exercise Print simulated arguments

Declare char *args[] = {"prog", "elma", "armut", "portakal"}; and int n = 4;. Loop with for (int i = 1; i < n; i++) and print each argument with printf("%s\n", args[i]);.

Show example solution

#include <stdio.h>

int main(void) {
    char *args[] = {"prog", "elma", "armut", "portakal"};
    int n = 4;
    for (int i = 1; i < n; i++) {
        printf("%s\n", args[i]);
    }
    return 0;
}

Test yourself

Match the loop parts

i = 1

Start value
i < argc

Continuation test
i++

Step increment
argv[i]

Current string
Why does the loop start at i = 1?
- Because loops must start at 1
- Because argc cannot be at least 1
- Because argv[0] is always NULL
- Because argv[0] is the program name
Show answer

Because argv[0] is the program name — argv[0] holds the program name; real user arguments start at argv[1].

Complete the loop condition

for (int i = 1; i < ____; i++)

Show answer

for (int i = 1; i < argc; i++)

Next lesson: getopt

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