Trouble inserting characters into a string

Hi friends,

I have an assignment in C where I have to insert a user inputted character into a string at the position the user states. I have the code below but it does not seem to work, let alone compile. I hate to come here for homework help, but I believe that its best learning by being taught by smart people, such as the users here! Anyways, the code I have so far is below.

Code:

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


void insert (char origString[], char insertion_char, int position_of_insertion)
{
    int new_length = strlen(origString) + 1;
    char new_str[new_length];
}


int main()
{
    char origString[20], insertion_char[1], new_str;
    int position_of_insertion, i, new_length;


    printf("Enter a string: ");
    scanf("%s", origString);




    printf("Enter a character to insert: ");
    scanf("%s", insertion_char);




    printf("Which position should the character be inserted before: ");
    scanf("%d", &position_of_insertion);


    for (i=0; i<position_of_insertion; i++)
    {
        new_str[i]= insert(origString, insertion_char, position_of_insertion);
    }


    for (i=position_of_insertion+1; i<new_length; i++)
    {
        new_str[i] = insert(origString, insertion_char, position_of_insertion);
    }


    printf("The new string is: %s", new_str);




    return 0;
}

The output should be like this:

Code:

Enter a String:lovecats4ever
Enter a character to insert: I
Which position should the character be inserted before:0
The new string is Ilovecats4ever

Have you tried to address the various warnings and errors?

Code:

main.c||In function 'insert':|
main.c|8|warning: unused variable 'new_str'|
main.c||In function 'main':|
main.c|36|error: subscripted value is neither array nor pointer|
main.c|36|warning: passing argument 2 of 'insert' makes integer from pointer without a cast|
main.c|5|note: expected 'char' but argument is of type 'char *'|
main.c|42|error: subscripted value is neither array nor pointer|
main.c|42|warning: passing argument 2 of 'insert' makes integer from pointer without a cast|
main.c|5|note: expected 'char' but argument is of type 'char *'|
main.c|46|warning: format '%s' expects type 'char *', but argument 2 has type 'int'|
||=== Build finished: 2 errors, 4 warnings ===|

You should code step by step - write one thing, compile, test, and verify - add some more code to bring you one step closer to your goal - compile, test, verify - repeat.
A development process

Replace

Code:

 char insertion_char[1]

to

Code:

 char insertion_char

since it is just a char.

Then of course replace this

Code:

printf("Enter a character to insert: ");
    scanf("%s", insertion_char);

with this

Code:

printf("Enter a character to insert: ");
    scanf("%c", &insertion_char);

Ok when a function returns nothing, then in her prototype there is as return value the keyword void.
As a result we can not assign what the function I just described to a variable.
If we want a function to return something, then we have to write the type of that something in the return value

e.g

Code:

int foo(void)
{
     a = 5;
     return 5;
}

int main(void)
{
      int b;
      b = foo();
  
      return 0;
}

If foo had a return value of void then we would have the compiler giving as an error!

At first make sure you understood what i said.

Are you sure the output should be like this?
I mean the letter in position zero is 'l' in terms of an array... For example in the same example you have, if I give position 3 what will happen?

If it was 3 it would be lovIecats4ever. Thats what I hoped the for loop was able to achieve.

So if the position is zero then it should be Iovecats4ever and not Ilovecats4ever .

Did you understand the rest I said about the code?

Yes I understand what you said about the code. Also no, I am inserting the character, not replacing. That is why it gets inserted into the string, instead of replacing the character.

Oh I see. Thank you. Ok so now apply what you learned from post #2 and #3 and post your updated code

Code:

 include <stdio.h># include <string.h>
int insert (char origString[], char insertion_char, int position_of_insertion)
{
    int new_length = strlen(origString) + 1;
    char new_str[new_length];


    return new_length;
}


int main()
{
    char origString[20], insertion_char, new_str;
    int position_of_insertion, i, new_length;


    printf("Enter a string: ");
    scanf("%s", origString);




    printf("Enter a character to insert: ");
    scanf("%c", &insertion_char);




    printf("Which position should the character be inserted before: ");
    scanf("%d", &position_of_insertion);


    for (i=0; i<position_of_insertion; i++)
    {
        new_str[i]= insert(origString, insertion_char, position_of_insertion);
    }


    for (i=position_of_insertion+1; i<new_length; i++)
    {
        new_str[i] = insert(origString, insertion_char, position_of_insertion);
    }


    printf("The new string is: %s", new_str);




    return 0;
}

I believe this is what you meant? Thanks for your help so far mate. I am getting 2 errors:

Code:

LINE 39 error: subscripted value is neither array nor pointer
LINE 44 error: subscripted value is neither array nor pointer

Tell us what you think that this line does?

Code:

new_str[i]= insert(origString, insertion_char, position_of_insertion);

And one more - What this one does

Code:

printf("The new string is: %s", new_str);

It runs the loop, stops at the user entered position point, and inserts the insertion_character. At least that was my intentions for both of them.

The value of new_str[i] becomes what the function "insert" is returning.

And the function "insert" does not change the size of any character arrays. Instead it creates a new array of size origString + 1 and then once the function has ended, it no longer exists. So if "hello" was in origString, the function would return 6. Also note that origString array was already strlen + 1 -> It has a '\0' character on the end.

new_str is also declared as a char -> Not a char* which you can malloc memory to and not an array.

Thanks guys for pointing me in the right direction. I figured out the problem!