Remove the last comma from c loop - c

Question

i have the belowo loop in c that print the prime number
for(int i = 2; i<=arraySize; i++)
{
//If arraySize is not 0 then it is prime
if (numbers[i]!=0)
printf("%d,",numbers[i]);
}
the out put after enter 50 for example is
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,
i want to to not print the last comma how i can do it
i tried this code but not working
printf("%d%s", numbers[i], (i !=arraySize-1) ? "," : "");

Answer 1

Instead of printing a comma after each number, print it before. Then you can use a variable to tell if this is the first number being printed, and not print the comma.
first=true;
for(int i = 2; i<=arraySize; i++)
{
//If numbers[i] is not 0 then it is prime
if (numbers[i]!=0) {
printf("%s%d", (first ? "" : ","), numbers[i]);
first = false;
}
}

Answer 2

I like both the other answers but just want to throw in this error prone variant on the same theme.
_Bool first = true;
for (int i = 2; i <= arraySize; i++) {
if (numbers[i] != 0) {
printf(",%d" + first, numbers[i]);
first = false;
}
}
If first is true the actual formatting string will become "%d". If it's false it'll become ",%d".

Answer 3

Simple: Use a pointer to the "prefix" string, printed AHEAD of the next value:
char *sep = "";
for(int i = 2; i <= arraySize; i++ ) {
if( numbers[i] ) {
printf( "%s%d", sep, numbers[i] );
sep = ", "; // I added a SP, too
}
}
Here's an alternative that uses a "limited scope" variable to index a static string. For clarity in this example, the array boundaries have been adjusted.
int main( void ) {
int numbers[] = { 1, 1, 4, 8, 9, 0, 7 };
int arraySize = sizeof numbers/sizeof numbers[0];
for( int i = 0, out = 0; i < arraySize; i++ )
if( numbers[i] )
printf("%s%d", &","[!out++], numbers[i] );
return 0;
}
The negated boolean post-incrementing value of out provides the address of the '\0' to the first instance, then the address of "," in subsequent instances.
1,1,4,8,9,7

Answer 4

The answers already here are fine, but I'd like to add a "simpler" solution. Simpler in that it doesn't require any further logic or extra variables. It does, however, require that you know that the first number is non-zero.
printf("%d", numbers[2]);
for (int i = 3; i < arraySize; i++)
{
if (numbers[i] != 0)
printf(",%d", numbers[i]);
}

Answer 5

I think the other answers overcomplicates things. I don't see any reason to have a test for every iteration in the loop. Instead, I'd simply do the special case first:
printf("%d", numbers[2]);
for (int i = 3; i <= arraySize; i++) {
if (numbers[i]!=0)
printf(",%d", numbers[i]);
}
This will however need some additional code to correctly handle the case where arraySize is lower than 3.
But I would choose another approach from the beginning, and that is writing a good function for printing an array. Could look like this:
void printArray(const int *array, int size) {
putchar('['); // Of course this is optional
if(size > 0) {
printf("%d", array[0]);
for(int i=1; i<size; i++)
printf(",%d", array[i]);
}
putchar(']'); // And this too
}
and then something like this:
int convertArray(const int *numbers, int *array, int size) {
int ret = 0;
for(int i=0; i<size; i++) {
if(number[i] != 0) {
array[ret] = numbers[i];
ret++;
}
}
return ret;
}