Question

So I am entering an array and a number which is the length of the array to a function called “Write” like this :

string write (int n, char t[100])

To keep it simple i would only like to write the values of the array out whit the help of this function like this:

{
    int i;

    for (i=1;i<=n;i++) {
        if (t[i]=='a') {
        printf("%c" , t[i]);
        }
    }

    return 0;
}

In the int main() only thing I did was I used a scanf for the n variable and gave from 1 to n a ‘a’ string value in the array.And I called the write function string write(n,t[100]); Here is the whole “main” :

    int main()  
{  
    int i,n;   
    char t[100];  
    scanf("%i" ,&n);  

    for (i=1;i<=n;i++) {
            t[i]='a';
            }
string write (n,t[100]);  
return 0;  
} 

My question is why wont this simple program run I can enter the value of the n but won’t print anything out? I must be missing something out whit the declaration’s or calling of the function I am new to C++.

Answer 1

string write(n, t[100]) is the prototype of your function (type omitted). If you want to call write, you need to do something like

int main()  
{  
    int i,n;   
    char t[100];  
    scanf("%i" ,&n);  

    for (i=1;i<=n;i++) {
            t[i]='a';
            }
write(n,t);  
return 0;  
}

Passing t[100] to your write function will give the 100th element of you string to your function, and that is not what you want to do. To pass the whole string, just use t.

Another mistake is that you say write returns a string. But you return 0 in your code, you want to modify the write prototype to int write(int n, char t[100]);