hey.
is this valid code?
void func1(int i, double j = 1.0) {
}
void main() {
auto fn = func1;
func1(1); //dmd: ok
fn(1); // dmd: not ok
}
On 08/22/2012 11:51 AM, Ellery Newcomer wrote:
hey.
is this valid code?
void func1(int i, double j = 1.0) {
}
void main() {
auto fn = func1;
func1(1); //dmd: ok
fn(1); // dmd: not ok
}
The type of the function pointer does not include the values of the
default parameters.
The type
On 08/22/2012 12:03 PM, Ali Çehreli wrote:
On 08/22/2012 11:51 AM, Ellery Newcomer wrote:
hey.
is this valid code?
void func1(int i, double j = 1.0) {
}
void main() {
auto fn = func1;
func1(1); //dmd: ok
fn(1); // dmd: not ok
}
The type of the function pointer does not
On Wednesday, August 22, 2012 11:51:45 Ellery Newcomer wrote:
hey.
is this valid code?
void func1(int i, double j = 1.0) {
}
void main() {
auto fn = func1;
func1(1); //dmd: ok
fn(1); // dmd: not ok
}
Default arguments are not part of the type. This behavior is very much on
On 08/22/2012 12:15 PM, Ellery Newcomer wrote:
On 08/22/2012 12:03 PM, Ali Çehreli wrote:
On 08/22/2012 11:51 AM, Ellery Newcomer wrote:
hey.
is this valid code?
void func1(int i, double j = 1.0) {
}
void main() {
auto fn = func1;
func1(1); //dmd: ok
fn(1); // dmd: not ok
}
The
