atoi() not working with char array

Question

I am learning C and I am trying to find how to much things with a database that I have made with char arrays.

I know atoi works with strings, but I cannot understand the difference between string and char array declaration (i understand strings have also a char with '/0').

It creates a warning when compiling [Warning] initialization makes integer from pointer without a cast at the line int w = LoggersID[j];

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

int main()
{
    // Logger ID Database
    char LoggersID[50][2];
    strcpy(LoggersID[1], "A");
    strcpy(LoggersID[2], "B");
    strcpy(LoggersID[3], "C");
    strcpy(LoggersID[4], "D");
    strcpy(LoggersID[5], "E");
    strcpy(LoggersID[6], "F");
    strcpy(LoggersID[7], "G");
    strcpy(LoggersID[8], "H");
    strcpy(LoggersID[9], "I");
    strcpy(LoggersID[10], "J");
    strcpy(LoggersID[11], "K");
    strcpy(LoggersID[12], "L");
    strcpy(LoggersID[13], "M");
    strcpy(LoggersID[14], "N");
    strcpy(LoggersID[15], "O");
    strcpy(LoggersID[16], "P");
    strcpy(LoggersID[17], "Q");
    strcpy(LoggersID[18], "R");
    strcpy(LoggersID[19], "S");
    strcpy(LoggersID[10], "T");
    strcpy(LoggersID[21], "1");
    strcpy(LoggersID[22], "2");
    strcpy(LoggersID[23], "3");
    strcpy(LoggersID[24], "4");
    strcpy(LoggersID[25], "5");
    strcpy(LoggersID[26], "6");
    strcpy(LoggersID[27], "7");
    strcpy(LoggersID[28], "8");
    strcpy(LoggersID[29], "9");
    strcpy(LoggersID[30], "10");


    printf("Lets start!\n");


    for (int i = 65; i < 86; i++)
    {
        for (int j = 1; j < 31; j++)
        {
            int w = atoi(LoggersID[j]);
            if (w == i)
            {
                printf("\nYou matched %s with %d", LoggersID[j], i);
            }
        }
    }

    for (int i = 1; i < 11; i++)
    {
        for (int j = 1; j < 31; j++)
        {
            int w = LoggersID[j];
            if (w == i)
            {
                printf("\nYou matched %s with %d", LoggersID[j], i);
            }
        }
    }
    printf("\nProgram finished!");
    getchar();
    return 0;
}

When I run it I get

Lets start!

Program finished!

instead of matching!

Answer 1

Your code has some problems:

• arrays are 0 based in C: the first element is at offset 0.

• the array elements should have at least 3 characters to accommodate for the null terminator for strings of 2 characters such as "10".

• you could use an initializer for the array instead of tediously initializing the elements with strcpy().

• converting the array elements from a string to a numeric value is done either by calling atoi() or strtol() to convert the string encoded as digits, or by reading the value of the first character with int w = LoggersId[i][0];. Converting the address of the element to an int is meaningless. The compiler issues a warning:

  [Warning] initialization makes integer from pointer without a cast
      at the line int w = LoggersID[j];
  

  Such warnings indicate programming errors and should be considered fatal. you can configure the compiler to help you avoid such mistakes with extra options on the command line, such as gcc -Wall -Werror or clang -Weverything -Werror.

Here is a modified version where you will see the matches for the 2 approaches:

#include <stdio.h>
#include <stdlib.h>

int main(void) {
    // Logger ID Database
    char LoggersID[30][3] = {
        "A", "B", "C", "D", "E", "F", "G", "H", "I", "J",
        "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T",
        "1", "2", "3", "4", "5", "6", "7", "8", "9", "10",
    };

    printf("Lets start!\n");
    for (int i = 1; i < 11; i++) {
        for (int j = 0; j < 30; j++) {
            int w = atoi(LoggersID[j]);
            if (w == i) {
                printf("You matched LoggersID[%d] = \"%s\" with %d\n",
                       j, LoggersID[j], i);
            }
        }
    }

    for (int i = 65; i < 85; i++) {
        for (int j = 0; j < 30; j++) {
            int w = LoggersID[j][0];
            if (w == i) {
                printf("You matched LoggersID[%d] = \"%s\" with %d (%c)\n",
                       j, LoggersID[j], i, i);
            }
        }
    }
    printf("\nProgram finished!");
    getchar();
    return 0;
}