Your problem is with this line:
else if (argc == 3 && argc == testscore) {In fact, when argc == 3, then you want to check if argv[2] is a numeric argument.
else if ( (argc==3) && (is_numeric_arg(argv[2])==1)) {A possible implementation would be:
int is_numeric_arg(char* arg) { int isInt = 0; int isFloat = 0; int isChar = 0; char* currChar; int i = 0; currChar = arg; isInt = 1; while (*currChar != '\0') { if(*currChar < '0' || *currChar > '9') { if(*currChar == '.'&& isInt == 1) { isInt = 0; isFloat = 1; } else { isInt = 0; isChar = 1; } } currChar++; } if (isChar == 1){ return 0; } // argument is a string else { return 1; } // argument is a int or float }int main (int argc, char *argv[]) { double testscore; if (argc == 2) { printf("Hello, Mr.%s.\n", argv[1]); } else if ( (argc==3) && (is_numeric_arg(argv[2])==1)) { testscore = atof(argv[2]); printf("Hi, Mr.%s, your score is %.1f\n", argv[1], testscore); } else { printf("My name is %s %s.\n", argv[1], argv[2]); }}I did not test the code and there is probably a better way to check that the argument from the command line is "numeric".
