Is it possible to declare auto variables with an if? - c++

Question

My Code is below.
struct conv{
struct des
{
des(int a) {}
des(int a, int b) {}
des(int a, int b, int c) {}
};
};
int main(int argc, const char * argv[]) {
int a = 1;
int b = 1;
int c = 1;
if (a > 1)
{
auto conv_desc = conv::des(1, 2, 3);
} else if(b > 1) {
auto conv_desc = conv::des(1, 2);
} else {
auto conv_desc = conv::des(1);
}
return 0;
}
The code's pattern is extracted from mkldnn. The only thing I want to do is take auto conv_desc out of the if-else statement.
I tried to declare auto conv_desc out of the if-else statement.
It occurred an error:
Declaration of
variable 'conv_desc' with deduced type 'auto' requires an initializer
Or if I used the pointer like below, I got a null pointer.
Another way got an error:
Taking the address of a temporary object of type 'conv::des'
If I can't solve this problem, I will have to write a large piece of duplicate code in each branch.

Answer 1

Move your if code into separate function:
conv::des make_des(int a, int b) {
if (a > 1) {
return conv::des(1, 2, 3);
} else if(b > 1) {
return conv::des(1, 2);
} else {
return conv::des(1);
}
}
int main(int argc, const char * argv[]) {
int a = 1;
int b = 1;
int c = 1;
auto conv_desc = make_des(a, b);
return 0;
}

Answer 2

Don't use auto. If you need to declare a variable and can't assign a value yet, then auto can't be used.
int main(int argc, const char * argv[]) {
// ...
conv::des conv_desc; // calls default constructor.
if (a > 1)
{
conv_desc = conv::des(1, 2, 3);
} else if(b > 1) {
conv_desc = conv::des(1, 2);
} else {
conv_desc = conv::des(1);
}
// conv_desc is initialized at this point.
return 0;
}
You can't use a pointer like this
int *a;
{ // new scope
int b = 5;
a = &b; // error.
}
// b is no longer in scope here
When b goes out of scope, then a will be pointing at the address where b used to be, which is now no longer valid.
If you want to use a pointer then you can use new. However in that case you must release the memory afterwards.
int main(int argc, const char** argv)
{
conv::des *conv_desc = nullptr; // does not call default constructor.
if (a > 1)
{
conv_desc = new conv::des(1, 2, 3);
} else if(b > 1) {
conv_desc = new conv::des(1, 2);
} else {
conv_desc = new conv::des(1);
}
if (conv_desc == nullptr) { /* memory allocation failed */ }
// conv_desc is initialized at this point.
// ...
// remember to delete conv_desc
if (conv_desc != nullptr) { delete conv_desc; conv_desc = nullptr; }
return 0;
}

Answer 3

You also can use an immediately called lambda. This is a common pattern to make some variables const when they could not otherwise.
This solution is very similar to #yachoor answer but uses a lambda.
int main(int argc, const char * argv[]) {
int a = 1;
int b = 1;
int c = 1;
// Kind of an inline function that is called immediately
auto const conv_desc = [&]{
if (a > 1) {
return conv::des(1, 2, 3);
} else if(b > 1) {
return conv::des(1, 2);
} else {
return conv::des(1);
}
}();
// Parens calls the function
return 0;
}

Answer 4

Seems like you are making the code overly-complicated. It is not clear why you would want to differentiate between these three constructors in this way. Why not hide the complexity inside the object by using default values like this?
struct conv{
struct des
{
des(int a, int b = 0, int c = 0) {
if(a > 1) {
/// do some logic
} else if(b > 1) {
// do some logic
} else {
// do some logic
}
}
};
};
int main(int argc, const char * argv[]) {
int a = 1;
int b = 1;
int c = 1;
auto conv_desc = conv::des(a, b, c);
return 0;
}

Answer 5

If you move the if ... else chain into a separate function, you can use return
struct conv
{
struct des
{
des(int a) {
}
des(int a, int b) {
}
des(int a, int b, int c) {
}
};
};
conv::des make_conv_des(int a, int b)
{
if (a > 1) {
return conv::des(1, 2, 3);
} else if(b > 1) {
return conv::des(1, 2);
} else {
return conv::des(1);
}
}
int main(int argc, const char * argv[]) {
int a = 1;
int b = 1;
auto conv_des = make_conv_des(a, b);
return 0;
}