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How can a function return a pointer to its calling routine?
The general form of a function is:
type_specifier function_name(parameter list)
{
body of function;
}
where type_specifier specifies the type of value that the function's return statement returns and the parameter list is a comma-separated list of variable names and types which receive the values of the arguments passed by the call to the function. The purpose of most functions is to either perform a computation and return a value, manipulate information and return a success-or-failure (true or false) value, or perform a strictly procedural routine which produces no value (eg, the exit() function). Functions may be declared to return any valid C data type. If the return type is not specified, it automatically defaults to type int. return causes an immediate exit from a function. It may also be used to return a value. All functions, except those of type void, return a value.
Functions that return pointers are handled just like any other type of function, e.g.:
char * match(char c, char *s)
while(c != *s && *s)
s++;
return(s);
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A using declaration in C++ makes it likely to use a name from a namespace without the scope operator.
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