You can use sprintf to do it, or maybe snprintf if you have it:
char str[ENOUGH];
sprintf(str, "%d", 42);
Where the number of characters (plus terminating char) in the str can be calculated using:
(int)((ceil(log10(num))+1)*sizeof(char))
You can use sprintf to do it, or maybe snprintf if you have it:
char str[ENOUGH];
sprintf(str, "%d", 42);
Where the number of characters (plus terminating char) in the str can be calculated using:
(int)((ceil(log10(num))+1)*sizeof(char))
As pointed out in a comment, itoa() is not a standard, so better use the sprintf() approach suggested in the rival answer!
You can use the itoa() function to convert your integer value to a string.
Here is an example:
int num = 321;
char snum[5];
// Convert 123 to string [buf]
itoa(num, snum, 10);
// Print our string
printf("%s\n", snum);
If you want to output your structure into a file there isn't any need to convert any value beforehand. You can just use the printf format specification to indicate how to output your values and use any of the operators from printf family to output your data.
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Simplest way to convert characters into a string? (C)
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Use sprintf():
int someInt = 368;
char str[12];
sprintf(str, "%d", someInt);
All numbers that are representable by int will fit in a 12-char-array without overflow, unless your compiler is somehow using more than 32-bits for int. When using numbers with greater bitsize, e.g. long with most 64-bit compilers, you need to increase the array size—at least 21 characters for 64-bit types.
Making your own itoa is also easy, try this :
char* itoa(int i, char b[]){
char const digit[] = "0123456789";
char* p = b;
if(i<0){
*p++ = '-';
i *= -1;
}
int shifter = i;
do{ //Move to where representation ends
++p;
shifter = shifter/10;
}while(shifter);
*p = '\0';
do{ //Move back, inserting digits as u go
*--p = digit[i%10];
i = i/10;
}while(i);
return b;
}
or use the standard sprintf() function.
But not using ascii. if i give "56", it has to return 56 in int format
Hello everyone!
In this simple example, how can I replace the new_key string contents with the uppercase of the key string? I can print it alright with the below code, but can't find a way to store the characters in the empty string - there are always either segmentation fault errors or initializer ones.
What would be the best way to make new_key[i] be the content of toupper(key[i])?
#include <stdio.h>
#include <cs50.h>
#include <string.h>
#include <stdlib.h>
#include <ctype.h>
int main(void)
{
string key = "abcde";
string new_key[] = "";
for (int i = 0, len = strlen(key); i < len; i++)
{
char c = key[i];
printf("%c", toupper(c));
}
}
I think the code below will do your work. I uses the standard sprintf function, which prints data from any type to a string, instead to stdout. Code:
#include <stdio.h>
#define INT_FORMAT "%d"
#define FLOAT_FORMAT "%f"
#define DOUBLE_FORMAT "%lf"
#define LL_FORMAT "%lld"
// ect ...
#define CONVERT_TO_STRING(FORMAT, VARIABLE, LOCATION) \
do { \
sprintf(LOCATION, FORMAT, VARIABLE); \
} while(false)
int main() {
char from_int[30];
char from_float[30];
char from_double[30];
char from_ll[30];
CONVERT_TO_STRING(INT_FORMAT, 34, from_int);
CONVERT_TO_STRING(FLOAT_FORMAT, 34.234, from_float);
CONVERT_TO_STRING(DOUBLE_FORMAT, 3.14159265, from_double);
CONVERT_TO_STRING(LL_FORMAT, 9093042143018LL, from_ll);
puts(from_int);
puts(from_float);
puts(from_double);
puts(from_ll);
return 0;
}
You will get a string version of the variable's name, i.e. it will convert a to "a". The #when used like this is called the stringification operator.
For example:
#define TO_STRING(val) #val
int main(void)
{
const int a = 12;
print("a is %s\n", TO_STRING(a));
return 0;
}
This will print a is a.
What do you expect to happen?
You can't get the variable's value, of course, since that's not available when the pre-processor runs (which is at compile-time).
To answer the question without reading too much else into it I would
Copychar 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.
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:
Copy...
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.