Reputation: 303
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?
Upvotes: 22
Views: 197478
Reputation: 16540
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); //
Upvotes: 0
Reputation: 121407
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);
Upvotes: 3
Reputation: 149
In C/C++ a string is an array of char terminated with a NULL byte ('\0'
);
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
Upvotes: 1
Reputation: 33904
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);
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.
Upvotes: 41
Reputation: 1079
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] = ".";
Upvotes: 0