CODEWITHSUNDEEP
arraysc
user14416725
user14416725

Reputation:

Program can't print entire array of strings in C but it either prints nothing or only the first character. What is wrong?

I want this array to print all the elements in this code but it doesn't.

#include <stdio.h>
#include <stdlib.h>
//#include "student_info.h"
int main()
{
    char animals[3];
    animals[0] = "lion";
    animals[1] = "Cat";
    numbers[2] = "Tiger";
    printf("%s", numbers[2]);
}

Upvotes: 0

Views: 369

Answers (2)

Tarik
Tarik

Reputation: 11209

You should have declared an array of array of char. What you have is an array of char that hold an individual character each.

const char *animals[3];
animals[0] = "Lion";
animals[1] = "Tiger";
animals[2] = "Wolf";

Another way to do it is:

char animals[3][10] = {
                     "Lion",
                     "Tiger",
                     "Wolf"
                 };

Working demo of both approaches:

#include <stdio.h>

int main() {
    const char *animals1[3];
    animals1[0] = "Lion";
    animals1[1] = "Tiger";
    animals1[2] = "Wolf";
    
    for (int i=0; i<3; i++)
        printf("%s\n", animals1[i]);


    char animals2[3][10] = {
                     "Lion",
                     "Tiger",
                     "Wolf"
                 };
                 
    for (int i=0; i<3; i++)
        printf("%s\n", animals2[i]);
                 
}

See Online Demo

Upvotes: 1

Yunnosch
Yunnosch

Reputation: 26763

Assuming numbers -> animals....

You declare an array of three single chars. char animals[3];

Then you attempt to write "strings" (there actually is no such thing in C) to places where only single chars are possible. animals[0] = "lion";
And you should get lots of warnings like

main.c:7:16: warning: assignment makes integer from pointer without a cast [-Wint-conversion]
     animals[0] = "lion";

Reading compiler warnings closely and treating them as errors is a habit I recommend adopting.

This one means that "lion" is a pointer to an (unchangeable) 0-terminated sequence of chars in memory. And you are trying to write that pointer into basically a char variable, where char is one of the types which for the compiler is a narrow integer.

Your code makes some sense if you change your array to store pointer to characters.

char* animals[3];

That matches how you init it and how you print it, with "%s" which is for pointers to 0-terminated sequences of characters, the closest thing C has to strings.

That code (remember the numbers...) gets you no warnings here
https://www.tutorialspoint.com/compile_c_online.php
and an output of:

Tiger

For your goal to print all entries in the array you need to understand that C does not have inbuilt support of printf() for arrys of "strings" (it "barely" manages char-sequences...).
So you will have to implement a loop yourself. That also gives you control over how it prints, one string per line, all in one line, what in between...

Upvotes: 2

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