Convert double value to a char array in C

If you are about to store the double DATA, not the readable representation, then:

#include <string.h> double a=2.132; char arr[sizeof(a)]; memcpy(arr,&a,sizeof(a)); 

To make a character string representing the number in human-readable form, use snprintf(), like in code below.

To access the bytes of the double, use a union. For example, union u { double d; char arr[8]; }

However, from your added comment, perhaps you mean to convert a number from characters to double. See the atof() call in code below. The code produces the following 4 output lines:

u.d = 2.132000 u.arr = 75 ffffff93 18 04 56 0e 01 40 res = 2.13200000 u.d = 37.456700 u.arr = ffffffa6 0a 46 25 75 ffffffba 42 40 res = 37.45670000 

Code:

#include <stdio.h> #include <stdlib.h> union MyUnion { char arr[8]; double d; }; void printUnion (union MyUnion u) { int i; enum { LEN=40 }; char res[LEN]; printf ("u.d = %f u.arr = ", u.d); for (i=0; i<8; ++i) printf (" %02x", u.arr[i]); printf ("\n"); snprintf (res, LEN, "%4.8f", u.d); printf ("res = %s\n", res); } int main(void) { union MyUnion mu = { .d=2.132 }; printUnion (mu); mu.d = atof ("37.4567"); printUnion (mu); return 0; } 

In case someone looks at this, you can also try sprintf:

Example:

char charray[200]; double num = 11.1111111111111; sprintf(charray, "%2.13f", num); 

Although I see some answers, I imagine you'd like to see code -- I'll just use snprintf although you might prefer a more secure form:

snprintf(arr, 8, "%2.4f", a); 

more here: http://msdn.microsoft.com/en-us/library/2ts7cx93(VS.71).aspx

If what you are asking is how to find out what bytes make up the double value in memory, try this:

double a=2.132; double darr[1] = { a }; char *arr = (char*) darr; 

although you probably want unsigned char, not char

double b=222.215;

 String bb=""+b; char[] tab=new char[bb.length()]; for(int m=0;m<bb.length();m++) { tab[i]=bb.charAt(m); }//un tableau mtn qui contient notre resultat dans un tableua de char