I cannot use the value passed in argv in the program call

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When I type b in the call of the program he does not return me the Bom dia. I would like to understand what is wrong in my code. Follow the program below.

#include <stdio.h>


int main (int argc, char *argv []){
  
  if (argv[1] == 'b'){
      printf("Bom dia");
  }
}
c c++

  • if (*argv[1] == 'b'){

2 answers

6


The parameter argv is a array of arrays of chars. When you use argv[1], what you’re actually accessing is a pointer to the start of array of chars which represents the second argument passed to the program.

On the line argv[1] == 'b', as argv[1] and 'b' are of different types the compiler makes a cast implicit variables. In this case what is compared is value of argv[1], a memory address, with 98, which is the ASCII value of the character 'b', as the two are always different the comparison fails.

Also know that although the following Compile code, it may not produce what you expect.

char str1[] = "abc";
char str2[] = "abc";

if (str1 == str2) {
    printf("iguais");
} else {
    printf("diferentes");
}

When you only access the name of array what you receive is the pointer to the first element of it, so what is compared are the memory addresses, not the contents of the strings.

For you to compare strings, which in C are only arrays of chars ended in NULL, you need to use the function strcmp header string.h

This function receives two strings and returns:

  • < 0 if str1 is less than str2
  • > 0 if str1 is greater than str2
  • 0 if str1 and str2 are the same

Your corrected code would look like this:

#include <stdio.h>
#include <string.h>

int main (int argc, char *argv []){
  if (argc < 2) return 0;
  if (strcmp(argv[1], "b") == 0){
      printf("Bom dia");
  }
}

Note that you should always check whether the correct number of arguments were passed to the program before accessing some value from argv, otherwise your program may access memory junk, which may produce unexpected results.

  • 1

    Hello, thank you very much for the answer, it was very helpful. I was able to understand perfectly where I was going wrong and I was able to learn a good lesson from it. Now, would you like to explain to me why when I run the program without passing any parameters it returns to me Segmentation fault? Thanks in advance.

  • 2

    @kodakz As I said in the reply, you should always check whether the number of arguments is correct according to the expected number. The argument argc saves the number of arguments passed to the program. You can place a block if at the beginning of the function main that makes this check: if (argc < 2) return 0; so the program ends running smoothly. I edited the answer to include this excerpt.

2

argv is an arrangement of pointers.

Better write:

#include <stdio.h>

int main (int argc, char *argv []){
    if (argc < 2) return 0;
    if (*argv[1] == 'b'){
        printf("Bom dia");
    }
}

or

#include <stdio.h>

int main (int argc, char *argv []){
    if (argc < 2) return 0;
    if (argv[1][0] == 'b'){
        printf("Bom dia");
    }
}

  • Grateful for the response, very good to see different ways to run the same program.

  • @kodakz, Thank you!

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