Type of array and printf specifiers

User 1144427

Hi there, I can't understand why my char type array isn't displayed with printf using string specifier %s. Here is my code.

#include
#include
#include

#define SIZE 10

void findbinary(int number, char result[], int index);

int main (void)
{
int someNumber = 233;
char result[SIZE];
int index = 0;
size_t i;
findbinary(someNumber, result,0);
printf("Decimal %d in binary = %c ",someNumber);
for(i = 0; result[i] != '\0'; i++)
printf("%s",result[i]);
}

void findbinary(int number, char result[], int index)
{

if(number == 0){
    return;
}
result\[index\] = number % 2;
findbinary(number / 2, result, index + 1);

}

It displayed properly if I use %d specifier, but... this is char array..? Thank you.

Mircea Neacsu

In:

printf("%s",result[i])

result[i] is a char not a string. It should be printed with '%c' not '%s'. Also in:

printf("Decimal %d in binary = %c ",someNumber);

you have 2 format specifiers (%d and %c) and only one argument. All hell can break loose when program accesses nonexistent argument.

Mircea

k5054

Your compiler will warn you about issues with printf formats not agreeing with the arguments provided. MS C seems to do this even without any additional warning flags. If you're using GCC (linux) or clang (Apple?) then you can add -Wall to the command line. The -Wall flag will generate warnings for the most often occurring code issues that are usually the cause of bugs. It's probably a good idea to add -Wextra to the command line, too.

Keep Calm and Carry On

CPallini

Probably you meant something similar to

#include #include #include #define SIZE 10

void findbinary(int number, char result[], int index);

int main (void)
{
int someNumber = 233;
char result[SIZE];
findbinary(someNumber, result,0);
printf("Decimal %d in reversed binary %s\n", someNumber, result);
return 0;
}

void findbinary(int number, char result[], int index)
{

if ( index == SIZE)
  exit(-1); //TODO: notify the error

if(number == 0){
    result\[index\] = '\\0'; // append theterminator
    return;
}
result\[index\] = (number % 2) + '0'; // obtain either the CHARACTER '0' or '1'
findbinary(number / 2, result, index + 1);

}

Note you are representing the binary number 'reversed' (that is leftmost bit is the least significant).

"In testa che avete, Signor di Ceprano?" -- Rigoletto