C String Formatting - c

Question

I wrote a program that is supposed to solve this this problem. "Write a program that, given a string, a width, and an empty string for output, centers the string in the output area. Use a function that returns 1 if the formatting is successful and 0 if any errors, such as string is greater then length". My problem is that the my program just returns a lot of odd characters when it prints the string out. And it won't flag 0. What can I do to fix my code and or solve the problem better?
FULL CODE:
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
int formatString(char str[250], char strcopy[250], int width);
int main()
{
char str[250];
char strcopy[250];
int width;
int outcome;
printf("Enter some text:\n");
gets(str);
printf("\nEnter a width (to check): ");
scanf("%d", &width);
printf("What you entered:");
printf("| %s |\n", str);
printf("\n");
outcome = formatString(str, strcopy, width);
if (outcome = 1)
{
printf("String copied.\n");
printf("| %s |", strcopy);
}
else
{
printf("Use a width val. that is the length of string\n");
}
return 0;
}
int formatString(char str[250], char strcopy[250], int width)
{
int sapceCheck;
int temp = 0;
sapceCheck = width - 1;
for (int i = 0; i < width; i++)
{
if (str[i] == '\0')
{
printf("Formating sucessful\n");
strncpy(str, strcopy, sizeof(str)-1); * (str + (sizeof(str) - 1)) = '\0';
temp = 1;
}
}
if (temp == 0)
{
return 0;
}
else
{
printf("Formating not sucessful\n");
printf("Width does not work\n");
return 0;
}
}

Answer 1

Don't miss the answer posted by 'Soravux', which contains all the right advice about how to fix the 'question' code.
Here is an alternate approach. However, the caller must ensure that the destination string 'strcpy' is large enough (length+1) to hold the output:
int formatString(char *str, char *strcopy, int length)
{
size_t strLength;
strLength = strlen(str);
/* Check if the string is greater than the length */
if(strLength > length)
return(0);
/* Print the centered 'str' to 'strcopy'. */
sprintf(strcopy, "%*s%*s",
(length+strLength) / 2, str, //leading spaces & 'str'
length - ((length+strLength) / 2), "" //trailing spaces
);
return(1);
}

Answer 2

A side note on your problem: returning 1 on success and 0 on failure is not the standard function behavior in C literature (aside from functions returning a boolean). Are you following a guide or a course?
You should not arbitrarily pre-allocate 250 chars for your strings. Strings are most often represented by char pointers, handled with dynamic memory allocation. I understand that gets() takes a pre-allocated buffer, but this is a really bad practice. You should consider using the fgets() function and use its maximum number of characters argument for a better security. Hard-coding string length is also hard to maintain (multiple places to modify the length) as well as confusing for readers (it's not clear if the compiler will pass the argument by copy or by reference).
Also, don't try to hide your code. The line containing the strncpy() have two statements, with the unary star pending in-between the two, looking like a multiplication operator. Don't do such things. Put the second statement on a new line.
That being said, you also misunderstood the arguments from the strncpy function: it is destination first, followed by source. You copy an uninitialized buffer (strcopy) into your string. Note that you also misinterpret the value of your temp variable which returns an error when something went well. It also returns 0 instead of 1 as asked.
Last notice: you really should break; your for-loop in your function formatString(), otherwise your strncpy will be executed for every \0 in your string.
I would recommend to check for exercises in a book if you are self-taught or seek the assistance of a TA if you are following a course. They will be able to guide you in the right path for understanding your mistakes.