A function that returns nth character from a string

Question

I'm trying to obtain the n-th character from a string. However, it's returning me the string from the n-th character onwards.

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

char *test(char input[],int position){

    char *result= malloc(sizeof(char)*100);
    result=&input[position];
    return result;
}

int main(){
    char *k;
    k=test("abcdefghi\n",3);
    printf("%s",k);
}

The above code returns defghi instead of just d. I'm not sure what's wrong here.

Answer 1

You're confusing strings and characters. You probably want this:

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

char test(char input[], int position){
   return input[position];
}

int main(){
    char k = test("abcdefghi\n",3);
    printf("%c", k);
}

This will print

d

This is so simple you don't actually need a function. Usually you have a pointer to a string or an array of characters and you just can pick the nth character like this:

char sometext[] = "abcdefghi\n";
...
char k = sometext[3];

Answer 2

You should probably just return a single char as shown in other answers.

But if you're really required to return char *, here's how you should do it.

You only need to allocate 2 characters -- one for the character you're returning, and one for the trailing null byte.

You need to copy the character into the allocated memory, not reassign the pointer variable.

char *test(char input[],int position){
    char *result= malloc(2);
    result[0] = input[position];
    result[1] = '\0';
    return result;
}

The caller should free the memory after printing it.

int main(){
    char *k;
    k=test("abcdefghi\n",3);
    printf("%s\n",k);
    free(k);
}

Answer 3

I'm trying to obtain the n-th character from a string. However, it's returning me the string from the n-th character onwards.

The function

char *test(char input[],int position){

does not return a character. It returns a pointer.

Moreover the function has a memory leak because at first there is allocated memory and its address is assigned to the pointer result and then the pointer is reassigned

char *result= malloc(sizeof(char)*100);
result=&input[position];

So the address of the allocated memory is lost.

Apart from this the parameter position can have a value that exceeds the length of the passed string. So the function can invoke undefined behavior.

If the function returns a pointer to a character then to output the pointed character you need 1) to dereference the pointer and 2) to use the conversion specifier %c instead of %s in a call of printf.

Also as the passed string is not being changed within the function then the corresponding parameter should be declared with the qualifier const.

The function can be declared and defined the following way

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

char * test( const char input[], size_t position )
{
    char *result = NULL;

    if ( position <= strlen( input ) )
    {
        result = ( char * )( input + position );
    }

    return result;
}

And in main you should write

char *k;

k = test( "abcdefghi\n", 3 );

if ( k != NULL ) printf( "%c\n", *k );

Answer 4

If you want to duplicate the string from the n-th position:

char *copyFronNthPosiztion(const char *input, size_t position)
{
    size_t len = strlen(input);
    char *result = NULL;
    if(position < len)
    {
        result = malloc(len - position + 1);
        strcpy(result, input + position);
    }
    return result;
}

If you want to have n-th character you do need any functions. Just use n-th index.

If you want to have a single char string:

#define NTH(str, pos)   ((char[]){(str)[pos], 0})

    k=NTH("abcdefghi\n",3);
    printf("%s",k);