c programming print ascii value of char*

Question

I'm trying to print the ascii value (ascii int numbers) of a char* for example A as 65 as on the ascii table .

Here is the code I have.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include <unistd.h>

int main(int argc, char *argv[])
{
    int i = 0;

    printf(" Hello World \n");
    printf("argv[0] is %s! \n",argv[0]);
    printf("argv[1] is %s! \n",argv[1]);
    printf("argv[2] is %s! \n",argv[2]);
    printf("argv[3] is %s! \n",argv[3]);

    if( argc == 4 )
    {
        printf("The first argument supplied is %s\n", argv[1]);
        printf("The second argument supplied is %s\n", argv[2]);
        printf("The third argument supplied is %s\n", argv[3]);
    }
    else if( argc > 4 )
    {
        printf("Too many arguments supplied.\n");
        exit( 1 );
    }
    else
    {
        printf("Not enough arguments supplied. \n");
        exit( 1 );
    }

    for(i = 1; i < 100; i++)
    {
        printf("i is %d argv[i] %d\n", i, argv[i]);
    }

    return(0);
}

I want these values to be entered from the command so this is what I put in.

./a.out A B C

When I compile it I get this warning.

warning: format '%d' expects type 'int', but argument 3 has type 'char *'

If I cast the argv[i] to (int) I get this message.

warning: cast from pointer to integer of different size

If I cast the argv[i] to (int *) I get this message.

warning: format '%d' expects type 'int', but argument 3 has type 'int *'

Answer 1

    printf("i is %d argv[i] %d\n", i, argv[i]);

is the problem causing line. Using incorrect format specifier is undefined behaviour in C.

argv is of type char** but %d format specifier expects an int . So to print the ascii values of arguments, you'll need a nested loop over argv .

    for(j=1;j<argc; j++) {
       for(i=0;argv[j][i];i++) {
          printf("i is %d argv[i] %d\n", j, argv[j][i]);
       }
    }

Note that your loop condition is not correct. You only have strlen(argv[j]) characters for each argument. So you can't arbitrarily print for 100 chars.

Move these printf statements

printf("argv[0] is %s! \n",argv[0]);
printf("argv[1] is %s! \n",argv[1]);
printf("argv[2] is %s! \n",argv[2]);
printf("argv[3] is %s! \n",argv[3]);

below the arguments checking so that you would only print them after arguments checking. Otherwise, you might end up accessing arguments that are not there if fewer arguments are passed.

Answer 2

• Point 1: Don't use argv[n] directly until you have checked for argc > n . This also includes unchecked loops like

   for(i = 1; i < 100; i++) 

  you'll end up accessing invalid memory after i gets equal to argc .

• Point 2: argv[n] is a null-terminated array or string (for valid cases), by default. You need %s format specifier to print that.

  That is, a command line A as the first argument to a.out is represented as "A" as argv[1] inside the program. It is not a char anymore, it is a string .