Convert Char to String in C - c

Question

How do I convert a character to a string in C. I'm currently using c = fgetc(fp) which returns a character. But I need a string to be used in strcpy

Answer 1

To answer the question without reading too much else into it I would
char str[2] = "\0"; /* gives {\0, \0} */
str[0] = fgetc(fp);
You could use the second line in a loop with whatever other string operations you want to keep using chars as strings.

Answer 2

You could do many of the given answers, but if you just want to do it to be able to use it with strcpy, then you could do the following:
...
strcpy( ... , (char[2]) { (char) c, '\0' } );
...
The (char[2]) { (char) c, '\0' } part will temporarily generate null-terminated string out of a character c.
This way you could avoid creating new variables for something that you already have in your hands, provided that you'll only need that single-character string just once.

Answer 3

Using fgetc(fp) only to be able to call strcpy(buffer,c); doesn't seem right.
You could simply build this buffer on your own:
char buffer[MAX_SIZE_OF_MY_BUFFER];
int i = 0;
char ch;
while (i < MAX_SIZE_OF_MY_BUFFER - 1 && (ch = fgetc(fp)) != EOF) {
buffer[i++] = ch;
}
buffer[i] = '\0'; // terminating character
Note that this relies on the fact that you will read less than MAX_SIZE_OF_MY_BUFFER characters

Answer 4

I use this to convert char to string (an example) :
char c = 'A';
char str1[2] = {c , '\0'};
char str2[5] = "";
strcpy(str2,str1);

Answer 5

A code like that should work:
int i = 0;
char string[256], c;
while(i < 256 - 1 && (c = fgetc(fp) != EOF)) //Keep space for the final \0
{
string[i++] = c;
}
string[i] = '\0';

Answer 6

//example
char character;//to be scanned
char merge[2];// this is just temporary array to merge with
merge[0] = character;
merge[1] = '\0';
//now you have changed it into a string

Answer 7

This is an old question, but I'd say none of the answers really fits the OP's question. All he wanted/needed to do is this:
char c = std::fgetc(fp);
std::strcpy(buffer, &c);
The relevant aspect here is the fact, that the second argument of strcpy() doesn't need to be a char array / c-string. In fact, none of the arguments is a char or char array at all. They are both char pointers:
strcpy(char* dest, const char* src);
dest : A non-const char pointer
Its value has to be the memory address of an element of a writable char array (with at least one more element after that).
src : A const char pointerIts value can be the address of a single char, or of an element in a char array. That array must contain the special character \0 within its remaining elements (starting with src), to mark the end of the c-string that should be copied.

Answer 8

Here is a working exemple :
printf("-%s-", (char[2]){'A', 0});
This will display -A-

Answer 9

FYI you dont have string datatype in C. Use array of characters to store the value and manipulate it. Change your variable c into an array of characters and use it inside a loop to get values.
char c[10];
int i=0;
while(i!=10)
{
c[i]=fgetc(fp);
i++;
}
The other way to do is to use pointers and allocate memory dynamically and assign values.