Incorrect output when printing an integer array

Question

I wanted to create a general print array function. This function would print array elements based on the data type.

#include "utilities.h"

//type : 0 ->int
//     : 1->char
void print_array(void *arr, int length, int type){
    int i = 0;

    for(i=0;i<length;i++){
        switch(type) {
            case 0:
                printf("Integer %d  %d\n", (int*)(arr+i));
                break;
            case 1:
               printf("%c \n", (char*)(arr+i));
               break;
            case 2:
                printf("%s \n", (char*)(arr+i));
                break;
            case 3:
                printf("%x \n", (int*)(arr+i));
                break;
            default:
                printf("Format not supported yet. %d \n",type);
                return;

        }
    }
}


void test_print_array(){
    int int_arr[3] = {0,1,2};
    print_array(int_arr,3,0);

}

int main(){
    test_print_array();
    return 0;
}

Output I am getting is

Integer -486474964

Integer -486474963

Integer -486474962

Instead of 0,1,2

I also tried using arr[i] instead of (int*)(arr+i) but I am getting compiler errors on that.

Post your compiler errors/warnings. Example: Review printf("%c \n", (char*)(arr+i));. %c expect a char` or int, not a pointer. — chux
– chux, Commented Apr 24, 2019 at 22:57
You're not dereferencing the pointers, and you're doing the pointer arithmetic on the uncast pointer and casting the result. Try eg. ((int *)arr)[i] instead of (int*)(arr+i) — Dmitri
– Dmitri, Commented Apr 24, 2019 at 22:58
You also might want to consider using an enum for the types... — Dmitri
– Dmitri, Commented Apr 24, 2019 at 23:00
You're using a compiler that allows pointer arithmetic on void*, like gcc. Those that do, consider the distance between "elements" to be 1. — molbdnilo
– molbdnilo, Commented Apr 24, 2019 at 23:18

Weather Vane · Accepted Answer · 2019-04-24 23:17:42Z

You are using pointer arithmetic on void* type, whose size is unknown. You must cast the pointer before doing the arithmetc, but it is clearer to use array notation anyway.

#include <stdio.h>

void print_array(void *arr, int length, int type){
    int i = 0;

    for(i=0;i<length;i++){
        switch(type) {
            case 0:
                printf("int %d\n", ((int*)arr)[i]);
                break;
            case 1:
                printf("char %c\n", ((char*)arr)[i]);
                break;
            case 2:
                printf("string %s \n", ((char**)arr)[i]);
                break;
            case 3:
                printf("hex %X\n", ((unsigned*)arr)[i]);
                break;
            default:
                printf("Format not supported yet. %d \n",type);
                return;

        }
    }
}

void test_int_array(){
    int int_arr[3] = {0, 1, 2};
    print_array(int_arr, 3, 0);

}

void test_char_array(){
    char char_arr[3] = {'x', 'y', 'z'};
    print_array(char_arr, 3, 1);

}

void test_str_array(){
    char *str_arr[3] = {"one", "two", "three"};
    print_array(str_arr, 3, 2);

}

void test_uns_array(){
    unsigned uns_arr[3] = {26, 27, 28};
    print_array(uns_arr, 3, 3);

}

int main(){
    test_int_array();
    test_char_array();
    test_str_array();
    test_uns_array();
    return 0;
}

Program output:

int 0
int 1
int 2
char x
char y
char z
string one 
string two 
string three 
hex 1A
hex 1B
hex 1C

chux · Accepted Answer · 2019-04-24 23:06:35Z

0

A general print array function is prone to undefined behavior (UB) unless the data reference is aligned properly and of a valid value.

Having said that, try

//printf("Integer %d  %d\n", (int*)(arr+i));
//              vv-------- print a `%`
printf("Integer %%d  %d\n", ((int*)arr)[i]);
//                          ^---------^ form a `int *` and then use [i]

Adding i to a void* is UB. Cast the void * pointer first to the desired pointer type, then use the index [i].

edited Apr 24, 2019 at 23:06

answered Apr 24, 2019 at 23:00

chux

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3 Comments

user968000 Over a year ago

utilities.c:11:47: error: operand of type 'void' where arithmetic or pointer type is required printf("Integer %d \n", (int*)arr[i]);

chux Over a year ago

@user968000 Your comment does not include the () of the answer. Use ((int*)arr)[i] to insure proper evaluation. (int*)arr[i] is like (int*)(arr[i]) which is invalid.

user968000 Over a year ago

works now : printf("Integer %d \n", ((int*)arr)[i]); thanks

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