в 14:21, 25 мая 2010

2010-05-25T14:21:17Z

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2010-05-25T10:29:44Z

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==Automatic type conversions can affect overloaded function resolution==

<source lang="cpp">#include <iostream>
using namespace std;

void f(int x);

void f(double x);

int main() {
int i = 1;
double d = 1.1;
short s = 9;
float r = 1.5F;

f(i); // calls f(int)
f(d); // calls f(double)

f(s); // calls f(int) -- type conversion
f(r); // calls f(double) -- type conversion

return 0;
}

void f(int x) {
cout << "Inside f(int): " << x << "\n";
}

void f(double x) {
cout << "Inside f(double): " << x << "\n";
}</source>
<pre class=codeResult>Inside f(int): 1
Inside f(double): 1.1
Inside f(int): 9
Inside f(double): 1.5</pre>

==Define overload function with short integer parameter==

<source lang="cpp">#include <iostream>
using namespace std;

void f(int x);
void f(short x);
void f(double x);

int main() {
int i = 1;
double d = 1.1;
short s = 9;
float r = 1.5F;

f(i); // calls f(int)
f(d); // calls f(double)

f(s); // now calls f(short)

f(r); // calls f(double) -- type conversion

return 0;
}

void f(int x) {
cout << "Inside f(int): " << x << "\n";
}

void f(short x) {
cout << "Inside f(short): " << x << "\n";
}

void f(double x) {
cout << "Inside f(double): " << x << "\n";
}</source>
<pre class=codeResult>Inside f(int): 1
Inside f(double): 1.1
Inside f(short): 9
Inside f(double): 1.5</pre>

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