How do i change std::cout
How do i change std::cout
#include <iostream>
#include <string>
using FText = std::string;
using Print = std::cout;
int main()
FText Submission1 = 0;
FText Submission2 = 0;
FText Operator = "";
return 0;
As you see I did,
using FText = std::string
how come i cant do a substitution for std::cout, The most confusing is the error i get
"No type named 'cout' in namespace 'std' "
can any pro coders help me im a begginer i dont like typing all if std::cout i managed to change string but cout is not working
im using xcode
cout
That sounds like a really bad idea.
– Danon
Aug 30 at 12:28
You can't write the code like this, for the same reason you can't write
int a; using TA = a;– Algirdas Preidžius
Aug 30 at 12:33
int a; using TA = a;
"change string", "change cout" These terms do not fit what you are doing, or are at best ambiguous. Please try to use full sentences to describe your goals. Cheers!
– Lightness Races in Orbit
Aug 30 at 12:43
Being lazy for typing is one big (if not the most frequent) cause for unneccessarily obfuscated code. Writing
std::cout is the most concice and readable way of writing std::cout, consider that there are millions of c++ programmers and every one of them knows what std::cout means, but only you know what Print is– user463035818
Aug 30 at 12:50
std::cout
std::cout
std::cout
Print
3 Answers
3
You can use using or typedef to introduce a new name for a type.std::string is a type, and these two are equivalent:
using
typedef
std::string
using FText = std::string;
typedef std::string FText;
But as the compiler says, std::cout is not a type (it's a variable).
std::cout
You can introduce a new name for std::cout using a reference:
std::cout
int main()
auto& Print = std::cout;
Print << "Hallo, Word!";
If you don't like look of namespace specifiers, you can do something like this:
int main()
using std::cout;
cout<< "Hallo, Word!";
You can create a new class say print class than override << insertion operator than use object of that class for printing.
class print
public:
int a;
void operator<<(print &obj)
cout<<obj.a<<endl;
;
print obj;
obj.a =1;
obj<<obj;
Required, but never shown
Required, but never shown
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coutis an instance, not a type.– πάντα ῥεῖ
Aug 30 at 12:27