In function 'print_usage': error: expected ')' before 'PROGRAM' - c

Question

As the title states, I am receiving an error when I try to build my project which states: In function 'print_usage': error: expected ')' before 'PROGRAM'
I am not exactly sure what is causing this error and was hoping someone could point me in the right direction.
#include <stdio.h>
#include <stdlib.h>
#include "lexer.h"
void print_usage();
void load_file(char *filename, char **buffer, size_t *size);
int main(int argc, char *argv[]) {
argc--;
argv++;
if(argc <= 0) {
print_usage();
exit(EXIT_FAILURE);
}
while(argc > 0) {
size_t size; /* I'm not using this, but it could be useful. */
Token token;
char *buffer = NULL;
unsigned int token_cnt = 0;
load_file(*argv, &buffer, &size);
if(buffer) {
char *b; /* editable copy of the buffer pointer */
b = buffer;
while(get_token(&b, &token) != TOKEN_EOF) {
printf("%s: ", id_string[token.id]);
if(token.str) {
printf("\"%s\"\n", token.str);
free(token.str);
}
else {
printf("no string found\n");
}
if(token.id != TOKEN_BAD) {
token_cnt++;
}
}
printf("\n----------------------\n");
printf("%u valid tokens found\n", token_cnt);
printf("----------------------\n");
free(buffer);
}
argc--;
argv++;
}
exit(EXIT_SUCCESS);
}
void print_usage() {
fprintf(stderr, "Usage: " PROGRAM " [file]...\n");
return;
}

Answer 1

argc--;
argv++;
/* Since you just incremented argv, you just lost the program name! */
/* You need argv[0] as the program name to create the error message */
if(argc <= 0) {
print_usage( argv[0] );
exit(EXIT_FAILURE);
}
void print_usage(char* program) {
fprintf(stderr, "Usage: %s [file]...\n", program);
}

Answer 2

Look up the syntax for fprintf(), you probably need something more like
void print_usage() {
fprintf(stderr, "Usage: %s [file]...\n", PROGRAM);
return;
}
Assuming PROGRAM is defined as a string somewhere.