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Here is a short program. It prints out the value of a variable "x". Ernie and Bert disagree about what will be printed: Ernie says, the value gets changed in "changeX" so it will print "7", and Bert says, no, when the function exits the changes get reversed and the value goes back to "5". Explain who is correct and why.
int x = 5;
void draw()
{
changeX();
print(x);
}
voidchangeX()
int x = 7;
decleration
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A: No. (Excluding if you have an old compiler, you might have to force the new operator to throw an exception if it runs out of memory.) This turns out to be a real pain to alwa
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Default argument: When the argument is missing then the function will read the default value of the missing argument. To make use of default argument functionality the argu
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