Printing Function in C++ not working

Question

Here is a code to print elements of an array. I am getting an error.Can anyone find out what is the problem with the code?

#include <iostream>

 using namespace std;

 int print_r(char arrName,int len){
            for(int i=0;i<len;i++){
                    cout<<arrName[i]<<"\t";        
            }
}

 int main(){

int a[3];
a[0]=1;
a[1]=2;
a[2]=3;

print_r(a,3);

return 0;
}

What do you *think* the problem is? What does the compiler *say* is the problem? Please exhibit *some* effort toward solving this yourself. Come on. — Rob Kennedy, Jan 03 '14 at 04:53
Why are you asking us? What problem do you get when you run the code? There is one obvious problem I can see. — Abhineet, Jan 03 '14 at 04:54
You still don't list what error you get. Have you *tried* to google the error you get? What information did that give you? — Sumurai8, Jan 03 '14 at 07:16

score 2 · Accepted Answer · answered Jan 03 '14 at 05:12

2

int print_r(char arrName,int len)

should be

void print_r(int *arrName,int len)

you are not retuning anything so it should be void, not int, and you declared your array as integer why do you use char?

answered Jan 03 '14 at 05:12

B.I.A

700
1
7
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Can I use it for both `int` and `char`,I think it is not possible I need another Function for `char array` like `void print_r(char *arrName,int len)` Can I make one function for both??? – Axeem Jan 03 '14 at 05:37
other than overloading i am not aware of another way of using the same function for different arguments, have look at this answer http://stackoverflow.com/questions/429125/override-and-overload-in-c – B.I.A Jan 03 '14 at 05:52
Use a template if you want to call the function with different types - `template void print_r(const Type* arrName, std::size_t len)` – Captain Obvlious Jan 03 '14 at 17:27

TryinHard · Answer 2 · 2014-01-03T06:48:47.610

It is an obvious error that you are passing int* to char argument in function. Hope this would help-

#include <iostream>

using namespace std;

void print_r(int arrName[],int len){
        for(int i=0;i<len;i++){
                cout<<arrName[i]<<"\t";        
        }
}

int main(){

int a[3];
a[0]=1;
a[1]=2;
a[2]=3;

print_r(a,3);

return 0;
}

instead you can also define prototype as

void print_r(int *arrName, int len)
{
 //Function body lies here
}

score 1 · Answer 3 · answered Jan 03 '14 at 05:18

Change::

int print_r(char arrName,int len){
     for(int i=0;i<len;i++){
     cout<<arrName[i]<<"\t";        
    }

To

int print_r(int* arrName,int len){ // You are passing int array to print_r
     for(int i=0;i<len;i++){
     cout<<arrName[i]<<"\t"; 
     return 0 ; // int print_r( int*, int ) will expect an int return value
    }

score 0 · Answer 4 · answered Jan 03 '14 at 05:12

When defining your function int print_r, change the return type to void , that is,

int print_r(char arrName,int len){
        for(int i=0;i<len;i++){
                cout<<arrName[i]<<"\t";        
        }
}

//Instead use this
void print_r(char arrName,int len){
        for(int i=0;i<len;i++){
                cout<<arrName[i]<<"\t";        
        }
}

Printing Function in C++ not working

4 Answers4