Rearrange the code to define a function "foo", which throws an exception with a value of "-100" if its parameter is greater than

Question

Rearrange the code to define a function "foo", which throws an exception with a value of "-100" if its parameter is greater than 999. Then "foo" catches its exceptions and prints "error!" to the screen.

void foo(int arg) {
try {
if (arg > 999) throw -100; }
catch (int x) {
cout << "error!" << endl; }
}

Answer 1

void foo(int arg) {
try {
if (arg > 999) throw -100; }
catch (int x) {
cout << "error!" << endl; }
}

Answer 2

I think the question is do my homework

Answer 3

What is your question?

Answer 4

catch (int) {
try {
}

Answer 5

void foo(int arg) {
try {
if (arg > 999) throw -100; }
catch (int x) {
cout << "error!" << endl; }
}

Answer 6

thanks all for your supports

Answer 7

Rearrange the code to define a function "foo", which throws an exception with a value of "-100" if its parameter is greater than 999. Then "foo" catches its exceptions and prints "error!" to the screen.

Answer:

void foo(int arg) {
try {
if (arg > 999) throw -100; }
catch (int x) {
cout << "error!" << endl; }
}
Q&A

Answer 8

thanks all