CODEWITHSUNDEEP
cascii
Mauro
Mauro

Reputation: 372

C How to convert ASCII numbers in an array to text

How do I convert this array with ASCII numbers to text, so every number get's a letter. So "72" gets H, 101 gets E and so on.

int a[] = {72, 101, 108, 108, 111, 32, 87, 111, 114, 108, 100};

Upvotes: 1

Views: 1362

Answers (4)

Subham Sharma
Subham Sharma

Reputation: 11

You can use array of structs since the array may contain different data type. Try this:

#include <stdio.h>
struct data
{ 
    int j; 
    char c;
};
int main() 
{
    int a[] = {72, 101, 108, 108, 111, 32, 87, 111, 114, 108, 100};
    int size = sizeof(a)/sizeof(a[0]);
    struct data value[size];
    for(int i=0;i<size;i++)
    {
        if(a[i]>=48 && a[i]<=57)
        {
            char ch= a[i];
            value[i].j =ch; 
        }
        else
        {
            char ch= a[i];
            value[i].c =ch;  
        }
    }
    for(int i=0;i<size;i++){
         if(a[i]>=48 && a[i]<=57)
            printf("%d",value[i].j); 
        printf("%c",value[i].c); 
         }
   return 0;
}

Output will be: Hello World (Hope it helps.)

Upvotes: 0

Lundin
Lundin

Reputation: 213701

Everything in C is numbers. The compiler can't tell a difference between the integer constant 72 and 'H', either results in a number and they even both have the type int. Characters and number formats only apply when you present raw data to a human user.

So there's no converting 72 to 'H', it's already in the correct format, the raw number 72. We simply need to tell the computer that we want this raw number to get printed as a character. The simplest way is printf("%c", ... but that one expects a parameter of type char. So we would have to convert the a[0] from int type to char, which we can do by means of a cast. printf("%c", (char)a[0]);.

Had the array been char a[] = {72, 101, 108, 108, 111, 32, 87, 111, 114, 108, 100}; then the cast wouldn't be necessary.

Upvotes: 1

KennetsuR
KennetsuR

Reputation: 788

  1. C Character Type C uses char type to store characters and letters. However, the char type is integer type because underneath C stores integer numbers instead of characters.
int main(int argc, char *argv[], char **env) {
    int a[] = {72, 101, 108, 108, 111, 32, 87, 111, 114, 108, 100};
    for (int i = 0; i < sizeof a / sizeof(int); ++i) {
        printf("%c", (char) a[i]);

    }
}

Upvotes: 2

aureliar
aureliar

Reputation: 1584

You can iterate over the array, and use the %c formating parameter to print the ascii characters of the int value

for (int i = 0 ; i < 11; i ++){
   printf("%c", a[i]);
}

Upvotes: 4

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