Concatenate string with error - c

Question

I have this code in C but I don't know why I get an error like Argument of type “char” is incompatible with parameter of type const char*
char number_string[size] = { NULL };
int counter = 0;
for (counter = 0; counter < strlen(input_string - 1); counter++)
{
temp = input_string[counter];
if (isdigit(temp))
{
strcat(number_string, temp); //temp variable has the error only in this line
}
}

Answer 1

You're mixing char and char *. The declaration of number_string[] might not warn you, because NULL may simply be defined as 0, which is legal as a char and as a pointer. But the variable temp is definately a problem: You don't show its declaration, but its first assignment makes it a char, and its second use in strcat assumes it is a char *.
If you want to add a single character at a time to a string, you'll have to do it by hand, something like this:
int nslen = strlen(number_string);
for ...
number_string[nslen++] = temp
number_string[nslen] = '\0';

Answer 2

char *strcat(char *dest, const char *src); expects its last argument to have type const char *
temp is a variable of type char.
So you're getting the error Argument of type “char” is incompatible with parameter of type const char*.
You could try something like this...
strncat(number_string, &temp, 1)
I usually favor something like...
sprintf (buffer, "%s%c", number_string, temp)

Answer 3

This is not an answer to your problem with types; it is a suggestion to tackle the problem in another way.
You want to fill a string with all numeric digits from the input string. There is no standard function to do this.
You could use strcat, but that function operates on strings, which must be zero-terminated. You could create a temporary string of two chars – one digit and one null-terminator – but that would be ineffective. strcat also requires you to ensure that you don't overflow the char buffer that you append to.
In cases like yours, it is usually easier to tackle the problem on a low level, where you create the char array yourself, one character at a time. For example, you can iterate through the input string with i and copy all digits to the number string by means of a second index, j:
#include <stdlib.h>
#include <stdio.h>
#include <ctype.h>
int main()
{
size_t size = 12;
char number[size];
const char *input = "Alpha 123/Bravo 456/Charlie 789/Delta 007";
int i, j;
j = 0;
for (i = 0; input[i] != '\0'; i++) {
if (isdigit((unsigned char) input[i]) && j + 1 < size) {
number[j++] = input[i];
}
}
number[j] = '\0';
puts(number);
return 0;
}
Note how the code keeps track of the characters in the number string and how it takes care not to overflow the buffer. The number string may be truncated, but it will always be null-terminated.
I've also used input[i] != '\0' to detect the end of the string (which is, by definition, the null terminator '\0') instead of calling strlen(input), which always starts looking for the terminator fro the beginning of the string.

Answer 4

There are multiple problems in your code:
calling strlen(input_string - 1) most likely invokes undefined behaviour. You probably meant strlen(input_string) - 1 which would still cause the loop to run too far if input_string is an empty string;
calling strlen for each iteration is inefficient anyway;
calling isdigit(temp) is incorrect if temp is a char and char is signed by default;
strcat cannot be used the way you call it;
you should check for potential buffer overflow if all digits do not fit in the destination array.
Here is a much simpler function to extract all digits from the input_string:
char number_string[size];
int i, j;
for (i = j = 0; input_string[i] != '\0'; i++) {
unsigned char uc = input_string[i];
if (j < size - 1 && isdigit(uc)) {
number_string[j++] = uc;
}
}
number_string[j] = '\0';