19
votes

How can I add '.' to the char Array := "Hello World" in C, so I get a char Array: "Hello World." The Question seems simple but I'm struggling.

Tried the following:

char str[1024];
char tmp = '.';

strcat(str, tmp);

But it does not work. It shows me the error: "passing argument 2 of ‘strcat’ makes pointer from integer without a cast" I know that in C a char can be cast as int aswell. Do I have to convert the tmp to an char array aswell or is there a better solution?

5
Str contains no string, just the size declaration heheJuan Bonnett
That was just a example with Hello World to describe the Problem. It must be empty as first in my real program. Program will fill it later. The problem just contains to add a char/int to an char Arraymissjohanna
@MikeCAT Wrong language.emlai

5 Answers

29
votes

strcat has the declaration:

char *strcat(char *dest, const char *src)

It expects 2 strings. While this compiles:

char str[1024] = "Hello World";
char tmp = '.';

strcat(str, tmp);

It will cause bad memory issues because strcat is looking for a null terminated cstring. You can do this:

char str[1024] = "Hello World";
char tmp[2] = ".";

strcat(str, tmp);

Live example.

If you really want to append a char you will need to make your own function. Something like this:

void append(char* s, char c) {
        int len = strlen(s);
        s[len] = c;
        s[len+1] = '\0';
}

append(str, tmp)

Of course you may also want to check your string size etc to make it memory safe.

2
votes

The error is due the fact that you are passing a wrong to strcat(). Look at strcat()'s prototype:

   char *strcat(char *dest, const char *src);

But you pass char as the second argument, which is obviously wrong.

Use snprintf() instead.

char str[1024] = "Hello World";
char tmp = '.';
size_t len = strlen(str);

snprintf(str + len, sizeof str - len, "%c", tmp);

As commented by OP:

That was just a example with Hello World to describe the Problem. It must be empty as first in my real program. Program will fill it later. The problem just contains to add a char/int to an char Array

In that case, snprintf() can handle it easily to "append" integer types to a char buffer too. The advantage of snprintf() is that it's more flexible to concatenate various types of data into a char buffer.

For example to concatenate a string, char and an int:

char str[1024];
ch tmp = '.';
int i = 5;

// Fill str here

snprintf(str + len, sizeof str - len, "%c%d", str, tmp, i);
1
votes

In C/C++ a string is an array of char terminated with a NULL byte ('\0');

  1. Your string str has not been initialized.
  2. You must concatenate strings and you are trying to concatenate a single char (without the null byte so it's not a string) to a string.

The code should look like this:

char str[1024] = "Hello World"; //this will add all characters and a NULL byte to the array
char tmp[2] = "."; //this is a string with the dot 
strcat(str, tmp);  //here you concatenate the two strings

Note that you can assign a string literal to an array only during its declaration.
For example the following code is not permitted:

char str[1024];
str = "Hello World"; //FORBIDDEN

and should be replaced with

char str[1024];
strcpy(str, "Hello World"); //here you copy "Hello World" inside the src array 
0
votes

I think you've forgotten initialize your string "str": You need initialize the string before using strcat. And also you need that tmp were a string, not a single char. Try change this:

char str[1024]; // Only declares size
char tmp = '.';

for

char str[1024] = "Hello World";  //Now you have "Hello World" in str
char tmp[2] = ".";
0
votes

Suggest replacing this:

char str[1024];
char tmp = '.';

strcat(str, tmp);

with this:

char str[1024] = {'\0'}; // set array to initial all NUL bytes
char tmp[] = "."; // create a string for the call to strcat()

strcat(str, tmp); //