Should I use goto statement here? - c++

Question

A fine gentleman told me that goto statements were bad, but I don't see how I can not use it here:
int main()
{
using namespace std;
int x;
int y;
int z;
int a;
int b;
Calc: //How can i get back here, without using goto?
{
cout << "To begin, type a number" << endl;
cin >> x;
cout << "Excellent!" << endl;
cout << "Now you need to type the second number" << endl;
cin >> y;
cout << "Excellent!" << endl;
cout << "Now, what do you want to do with these numbers?" << endl;
cout << "Alt. 1 +" << endl;
cout << "Alt. 2 -" << endl;
cout << "Alt. 3 *" << endl;
cout << "Alt. 4 /" << endl;
cin >> a;
if (a == 1) {
z = add(x, y);
}
if (a == 2) {
z = sub(x, y);
}
if (a == 3) {
z = mul(x, y);
}
if (a == 4) {
z = dis(x, y);
}
}
cout << "The answer to your math question is ";
cout << z << endl;
cout << "Do you want to enter another question?" << endl;
cout << "Type 1 for yes" << endl;
cout << "Type 0 for no" << endl;
cin >> b;
if (b == 1) {
goto Calc;
}
cout << "Happy trails!" << endl;
return 0;
}
It is a calculator, as you can see. Also, if you want, can you suggest a better way (If it exists) to let the user choose the operation (+ - * /). Header files are under control.
I apologize for a lot of cout statements.

Answer 1

Here is a cleaned-up and properly formatted version using a do/while loop for structure:
using namespace std;
int main()
{
int x, y, z, a, b;
do {
cout << "To begin, type a number" << endl;
cin >> x;
cout << "Excellent!" << endl;
cout << "Now you need to type the second number" << endl;
cin >> y;
cout << "Excellent!" << endl;
cout << "Now, what do you want to do with these numbers?" << endl;
cout << "Alt. 1 +" << endl;
cout << "Alt. 2 -" << endl;
cout << "Alt. 3 *" << endl;
cout << "Alt. 4 /" << endl;
cin >> a;
if (a == 1) {
z = add(x, y);
}
else if (a == 2) {
z = sub(x, y);
}
else if (a == 3) {
z = mul(x, y);
}
else if (a == 4) {
z = dis(x, y);
}
cout << "The answer to your math question is ";
cout << z << endl;
cout << "Do you want to enter another question?" << endl;
cout << "Type 1 for yes" << endl;
cout << "Type 0 for no" << endl;
cin >> b;
} while (b != 0);
cout << "Happy trails!" << endl;
return 0;
}

Answer 2

Erm , use a proper looping construct, while, for etc.

Answer 3

the "more generally accepted" approach in this case would be a do {...} while(b==1); but the compiled results would likely be identical.

Answer 4

goto makes it difficult to track where execution is coming from, and where it's going.
goto encourages spagetti-code, unless you restrict heavily where it is used (e.g. you could argue that you only use it for cleanup blocks, but such an argument makes no sense in the presence of RAII).
you are using a goto to simulate a loop. Why are you not writing a loop instead?
it's obscure and thus, makes your code less available to other people.
goto makes it more difficult to track objects lifetimes.

Answer 5

Short answer to actual question: No, you should not use goto in this code. There is no need for it.
The use of goto should be "when it makes the code clearer or safer". The typical example of "makes the code clearer" is when there several layers of nested loops, and some particular situation requires leaving all the nesting levels, and adding a "do we want to exit the loop" makes the code more complicated. An example of "making it safer" is if a function holds a lock, opens a file or something similar, and needs to return early - but you also need to close the file or release the lock, using "goto exit_now;" is safer than trying to remember what locks, files, etc are held and then doing return;.
This:
if (a == 1) {
z = add(x, y);
}
if (a == 2) {
z = sub(x, y);
}
if (a == 3) {
z = mul(x, y);
}
if (a == 4) {
z = dis(x, y);
}
is a classic case of you should use 'switch':
switch(a)
{
case 1:
z = add(x, y);
break;
case 2:
z = sub(x, y);
break;
....
}
Makes the code clearer - there is also no confusion about whether a changes value and maybe another if statement becomes viable.

Answer 6

You can easily avoid 'goto' in your code. Just divide it into functions:
using namespace std;
void question () {
cout << "To begin, type a number" << endl;
cin >> x;
// put rest of the code here
}
int main () {
int ask = 1;
while ( ask == 1 ) {
question();
cout << "Do you want to enter another question?" << endl;
cout << "Type 1 for yes" << endl;
cout << "Type 0 for no" << endl;
cin >> ask;
}
return 0;
}
Edit: as noted in the comments, using do-while would be actually an better option.

Answer 7

goto isn't automatically bad. Unreadable code is bad. Whenever you find yourself in need of some obscure programming construct like 'goto', that usually means that your code is either poorly written, or that your program design is flawed.
The solution is almost always more functions. For example:
bool run_program();
int prompt_user_begin();
int prompt_user_again();
int prompt_operation_type();
bool prompt_continue();
int main()
{
while(run_program())
{}
cout << "Happy trails!" << endl;
return 0;
}
bool run_program()
{
int first;
int second;
int operation_type;
int result;
first = prompt_user_begin();
cout << "Excellent!" << endl;
second = prompt_user_again();
cout << "Excellent!" << endl;
operation_type = prompt_operation_type();
switch(operation_type)
{
case 1: result = add(first, second); break;
case 2: result = sub(first, second); break;
case 3: result = mul(first, second); break;
case 4: result = div(first, second); break;
}
cout << "The answer to your math question is ";
cout << result << endl;
return prompt_continue();
}
int prompt_user_begin ()
{
int x;
cout << "To begin, type a number" << endl;
cin >> x;
return x;
}
int prompt_user_again ()
{
int x;
cout << "Now you need to type the second number" << endl;
cin >> x;
return x;
}
int prompt_operation_type ()
{
int x;
cout << "Now, what do you want to do with these numbers?" << endl;
cout << "Alt. 1 +" << endl;
cout << "Alt. 2 -" << endl;
cout << "Alt. 3 *" << endl;
cout << "Alt. 4 /" << endl;
cin >> x;
return x;
}
bool prompt_continue ()
{
int x;
cout << "Do you want to enter another question?" << endl;
cout << "Type 1 for yes" << endl;
cout << "Type 0 for no" << endl;
cin >> x;
return x==1;
}