"ld returned 1 exit status" in simple program - multiplication

Question

can anyone tell me what is wrong with this code?
Im trying to pass the value of 'amount' and 'count' into int main where it displays the multiplication. I get the error "C:\Users\XXXX\Desktop\collect2.exe [Error] ld returned 1 exit status" with the int main() highlighted. Thank you.
#include <stdio.h>
void show(int, int);
int display()
{
int amount;
int count;
amount = 10;
count = 20;
return(0);
}
int main(int amount, int count)
{
show(amount, count);
int sum;
sum = amount * count;
printf("%d", sum);
return(0);
}

Answer 1

The error's cause is that your main function's signature is rejected by ld, which is expected because the standard signatures for main are int main(void) and int main(int argc, char** argv).
If you want to pass amount and count as parameters to main, that is as command-line parameters, you should use the second signature I gave for main, where argv is an array of the command line parameters and argc gives the number of elements in argv. Note that argv[0] is not a command line parameter but the name under which your program was called and that argv[argc] is a null pointer.

Answer 2

You should pass command line parameters to the main using argc and *argv[] instead of trying to define your own.
int main(int argc, char *argv[]) {
for (int i = 0; i < argc; ++i) {
std::cout << argv[i] << std::endl;
}
}
This prints the paramters passed to main() via the command line.
If you pass a value for amount a and a value for count via the command line, you can access them like this
int main(int argc, char *argv[]) {
std::cout << "amount" << argv[1] << std::endl;
std::cout << "count" << argv[2] << std::endl;
}
Used index 1 and 2 of argv because index 0 contains the program name
You specified no language so I used C++ for the above example, but the idea is the same for C:
int main(int argc, char *argv[]) {
printf("amount %s", argv[1]);
printf("count %s", argv[2]);
}
Note that the incoming values are char arrays, not integers