Reputation: 477
After verifying that strcat is where the error occurs, I then check the previous example in my assignment. In my previous examples I use strcat(actually strncat) in the same fashion as I do for my following code. I am not too sure.
The purpose of my program is to loop through "string" and remove any occurances the character 'c' from string.
main.c:
char string[100]={0}, c[3];
printf("Enter a String: ");
fgets(string, 100, stdin);
if (string[98] == '\n' && string[99] == '\0') { while ( (ch = fgetc(stdin)) != EOF && ch != '\n'); }
printf("Enter a Char: ");
fgets(c, 2, stdin);
while ( (ch = fgetc(stdin)) != EOF && ch != '\n');
rmchr(string, c[0]);
header:
rmchr(char *string, char c)
{
int i=0;
char *word[100];
int s = strlen(string);
for(i=0; i<=(s-2); i++)
{
if(string[i] != c)
{
strcat(word, string[i]);
}
}
}
Upvotes: 0
Views: 172
Reputation: 3433
The following seems to compile fine but your code doesnt do quite what you said you wanted, "The purpose of my program is to loop through "string" and remove any occurances the character 'c' from string.". the function doesn't remove the character or return a copy of the string with the character excluded. I wrote a function that copies the string after removing the character and returns pointer to it. below is your code a bit modified and under it is my function
//Just a compilable version of your code, not sure if it does what u want
#include <stdio.h>
#include <string.h>
void rmchr(char *string, char c)
{
int i=0;
char word[100];
int s = (int)strlen(string);
for(i=0; i<=(s-2); i++)
{
if(string[i] != c)
{
strcat(word, (char *)(&string[i]));
}
}
}
int main(int argc, const char * argv[]) {
char string[100] = {0}, c[3];
char ch;
printf("Enter a String: ");
fgets(string, 100, stdin);
if (string[98] == '\n' && string[99] == '\0') {
while ( (ch = fgetc(stdin)) != EOF && ch != '\n');
}
printf("Enter a Char: ");
fgets(c, 2, stdin);
while ( (ch = fgetc(stdin)) != EOF && ch != '\n');
rmchr(string, c[0]);
return 0;
}
There you go, with a demo main
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char* rmchr(char *string, char ch)
{
int counter = 0;
int new_size = 0;
char corrected_string[100];
while (string[counter] != '\n' && string[counter] != '\0' && string[counter] != EOF) {
if (string[counter] != ch) {
corrected_string[new_size] = string[counter];
new_size++;
}
counter++;
}
char *new_string = (char *)malloc((new_size+1) * sizeof(char));
for (int j = 0; j <= new_size; j++) {
new_string[j] = corrected_string[j];
}
return new_string;
}
int main(int argc, const char * argv[]) {
char *s = "The char 'c' will be removed";
char *new = rmchr(s, 'c');
printf("%s", new);
return 0;
}
Upvotes: 0
Reputation: 134356
There are more than one point to mention here, like
rmchr()
definition should have a return type, maybe void
if you're not returning anything.
[FWIW, In that case, I wounder, how you'll make use of the local variable word
]
inside rmchr()
, word
needs to be an array of char
s, not char
pointers. You need to change char * word[100]
to char word[100]
.
In strcat()
, both the arguments, needs to be a pointer. You need to use &string[i]
, in that case.
Upvotes: 0
Reputation: 145899
char *word[100];
It will hold a string in your program so use:
char word[100];
that is, an array of char
instead of an array of char *
.
Then strcat
concatenates to a string but word
is not initialized. Make it a string with:
word[0] = '\0';
Then string[i]
is a character but strcat
needs pointers to character arguments: to use a pointer use &string[i]
.
Finally the problem in your rmchr
function is it has to return something, either through the arguments or via a return statement but it doesn't.
Upvotes: 2