Closed
Description
Consider the following code:
try {
xxxx;
} catch(e) {
print(e);
}
Will not print out the expected ReferenceError
string. But if we change the code in a way that the xxxx
variable is part of an argument list then the exception is correctly raised.
So using the following code:
try {
print(xxxx);
} catch(e) {
print(e);
}
Will print out the expected ReferenceError
string.
Jerry version:
Checked revision: ac87616f0
Build: debug.linux