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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 argument must be initialized in the prototype or declaration.
Rules for declaring default argument:
1. The default value must given from right to left arguments
2. Cannot assign only the first argument as default when there is more than one argument.
3. All the arguments may be default.
4. When there is more than one argument the intermediate argument cannot be a default.
int f_def(int x, int y=20); // Default argument value initialized at prototype.
void main( )
{int a=10, b=15,c;
c = f_def(a, b); //The value of c is 15 c = f_def(a); // The value of c is 30
}
int f_def(int x, int y)
{ return (x+y);
flow chart of volume of sphere
flow chart for push operation in stack
What if I wish a local to "die" before the close} of the scope wherein it was created? Can I call a destructor on a local if I want to?
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Here is a program that uses an inline function to compute and return the absolute value of its input argument. # include inline int abs(int x) { ret
Write a ''C'' program to accept any 3 digit integer number from the keyboard and display the word equivalent representation of the given number.
A constructor that accepts one argument of a dissimilar type.
A: no, can't. The names of, precedence of, associativity of, and arity of operators is fixed through the language. In C++, there is no operator** so you cannot create one for a
calculate the total resisitance of a series circuit?
if one won't explicitly call the destructor of a local; then how does he handle the above situation?
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