Question

I need to check if my program is correctly inputting these characters in my strings, so how can I see a “raw” string with these characters not parsed but actually shown?

Answer 1

If you read input (from file or from user) the special escape codes are not parsed. It’s only in string and character literals in the source that the compiler treats those specially.

Edit: Simple example program with input and output to show what I’m talking about.

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

int main(int ac, char *av[])
{
    char input[32];

    printf("Enter input: ");
    fgets(input, sizeof(input), stdin);

    /* Remove trailing newline */
    if (input[strlen(input) - 1] == '\n')
        input[strlen(input) - 1] = '\0';

    printf("input is \"%s\"\n", input);

    return 0;
}

Example run of the above program:

Enter input: foo\nbar\thello
input is "foo\nbar\thello"

The function fgets leaves the actual newline at the end of the string. However, the sequences \n and \t in the input does not get translated to newline or tab (respectively). That is because it’s not the input or output functions that handles these special character sequences, but the compiler.

If you have those sequences inside a string or character literal in the source, then the compiler recognizes those and changes them to a proper newline, tab or whatever it is you wrote. However, as the compiler doesn’t know anything about input read from a file or from a user, these sequences are not translated.

Edit 2: In case you wonder about how to show literal special characters in a string, then please see this program:

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

void print_raw_string(const char *str)
{
    while (*str != '\0')
    {
        if (isprint(*str))
            fputc(*str, stdout);
        else
        {
            switch (*str)
            {
                /* First check for known special sequences */
            case '\0':
                printf("\\0");
                break;
            case '\a':
                printf("\\a");
                break;
            case '\b':
                printf("\\b");
                break;
            case '\t':
                printf("\\t");
                break;
            case '\n':
                printf("\\n");
                break;
            case '\v':
                printf("\\v");
                break;
            case '\f':
                printf("\\f");
                break;
            case '\r':
                printf("\\r");
                break;

            default:
                /* None of the above, print it out as a hex escape sequence */
                printf("\\x%02x", *str);
                break;
            }
        }

        str++;
    }
}

int main(int ac, char *av[])
{
    char input[32];

    printf("Enter input: ");
    fgets(input, sizeof(input), stdin);

    printf("Input is: ");
    print_raw_string(input);
    printf("\n");

    return 0;
}

When running the program:

Enter input: foo    bar
Input is: foo\tbar\n