How to swap between functions implementation?
7
Is there any way to swap between two functions implementation in C++ ?
Something like this:
void printA(); // print a char
void printB(); // print b char
printA(); // output: a
printB(); // output: b
functionSwap(printA, printB);
printA(); // output: b
printB(); // output: a
I want to use it with the ExitProcess function.
c++ c++11
edited Nov 13 at 14:29
Matthieu Brucher
11.8k22138
asked Nov 13 at 14:24
Mor Ben-Yosef
517
add a comment |
7
Is there any way to swap between two functions implementation in C++ ?
Something like this:
void printA(); // print a char
void printB(); // print b char
printA(); // output: a
printB(); // output: b
functionSwap(printA, printB);
printA(); // output: b
printB(); // output: a
I want to use it with the ExitProcess function.
c++ c++11
edited Nov 13 at 14:29
Matthieu Brucher
11.8k22138
asked Nov 13 at 14:24
Mor Ben-Yosef
517
5
This seems to be a possible instance of the XY problem. What is the root goal you're trying to achieve by swappingExitProcess?
– Andrey Akhmetov
Nov 13 at 14:26
@AndreyAkhmetov I am participating in some kind of competition in which everyone want to make the enemy process stop and there is the option of using DLL injection and using the ExitProcess function in order to make my process exit so I want to avoid the option of others to use the function in my process.
– Mor Ben-Yosef
Nov 13 at 14:31
add a comment |
7
7
7
Is there any way to swap between two functions implementation in C++ ?
Something like this:
void printA(); // print a char
void printB(); // print b char
printA(); // output: a
printB(); // output: b
functionSwap(printA, printB);
printA(); // output: b
printB(); // output: a
I want to use it with the ExitProcess function.
c++ c++11
edited Nov 13 at 14:29
Matthieu Brucher
11.8k22138
asked Nov 13 at 14:24
Mor Ben-Yosef
517
Is there any way to swap between two functions implementation in C++ ?
Something like this:
void printA(); // print a char
void printB(); // print b char
printA(); // output: a
printB(); // output: b
functionSwap(printA, printB);
printA(); // output: b
printB(); // output: a
I want to use it with the ExitProcess function.
c++ c++11
c++ c++11
edited Nov 13 at 14:29
Matthieu Brucher
11.8k22138
asked Nov 13 at 14:24
Mor Ben-Yosef
517
edited Nov 13 at 14:29
Matthieu Brucher
11.8k22138
asked Nov 13 at 14:24
Mor Ben-Yosef
517
edited Nov 13 at 14:29
Matthieu Brucher
11.8k22138
edited Nov 13 at 14:29
Matthieu Brucher
11.8k22138
edited Nov 13 at 14:29
Matthieu Brucher
11.8k22138
11.8k22138
asked Nov 13 at 14:24
Mor Ben-Yosef
517
asked Nov 13 at 14:24
Mor Ben-Yosef
517
asked Nov 13 at 14:24
Mor Ben-Yosef
517
517
5
This seems to be a possible instance of the XY problem. What is the root goal you're trying to achieve by swappingExitProcess?
– Andrey Akhmetov
Nov 13 at 14:26
@AndreyAkhmetov I am participating in some kind of competition in which everyone want to make the enemy process stop and there is the option of using DLL injection and using the ExitProcess function in order to make my process exit so I want to avoid the option of others to use the function in my process.
– Mor Ben-Yosef
Nov 13 at 14:31
add a comment |
5
This seems to be a possible instance of the XY problem. What is the root goal you're trying to achieve by swappingExitProcess?
– Andrey Akhmetov
Nov 13 at 14:26
@AndreyAkhmetov I am participating in some kind of competition in which everyone want to make the enemy process stop and there is the option of using DLL injection and using the ExitProcess function in order to make my process exit so I want to avoid the option of others to use the function in my process.
– Mor Ben-Yosef
Nov 13 at 14:31
5
5
This seems to be a possible instance of the XY problem. What is the root goal you're trying to achieve by swapping
ExitProcess?– Andrey Akhmetov
Nov 13 at 14:26
This seems to be a possible instance of the XY problem. What is the root goal you're trying to achieve by swapping
ExitProcess?– Andrey Akhmetov
Nov 13 at 14:26
@AndreyAkhmetov I am participating in some kind of competition in which everyone want to make the enemy process stop and there is the option of using DLL injection and using the ExitProcess function in order to make my process exit so I want to avoid the option of others to use the function in my process.
– Mor Ben-Yosef
Nov 13 at 14:31
@AndreyAkhmetov I am participating in some kind of competition in which everyone want to make the enemy process stop and there is the option of using DLL injection and using the ExitProcess function in order to make my process exit so I want to avoid the option of others to use the function in my process.
– Mor Ben-Yosef
Nov 13 at 14:31
add a comment |
2 Answers
2
active
oldest
votes
16
You can bind a pointer to both functions in two variables and swap those.
void (*f1)() = printA;
void (*f2)() = printB;
f1(); // output: a
f2(); // output: b
std::swap(f1, f2);
f1(); // output: b
f2(); // output: a
answered Nov 13 at 14:28
lubgr
10.2k21745
Don't use c function pointers, butstd::functioninstead
– hellow
Nov 13 at 14:29
13
@hellow no.std::functionis a heavywieght type-erased container for callable objects. If a function pointer is enough, use that.
– Quentin
Nov 13 at 14:32
add a comment |
7
You need to wrap them in objects (or pointers to functions):
std::function<void()> myprintA = printA;
std::function<void()> myprintB = printB;
std::swap(myprintA, myprintB);
myprintA();
myprintB();
Otherwise, you are working with symbols themselves, and you can't swap this.
answered Nov 13 at 14:28
Matthieu Brucher
11.8k22138
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
16
You can bind a pointer to both functions in two variables and swap those.
void (*f1)() = printA;
void (*f2)() = printB;
f1(); // output: a
f2(); // output: b
std::swap(f1, f2);
f1(); // output: b
f2(); // output: a
answered Nov 13 at 14:28
lubgr
10.2k21745
Don't use c function pointers, butstd::functioninstead
– hellow
Nov 13 at 14:29
13
@hellow no.std::functionis a heavywieght type-erased container for callable objects. If a function pointer is enough, use that.
– Quentin
Nov 13 at 14:32
add a comment |
16
You can bind a pointer to both functions in two variables and swap those.
void (*f1)() = printA;
void (*f2)() = printB;
f1(); // output: a
f2(); // output: b
std::swap(f1, f2);
f1(); // output: b
f2(); // output: a
answered Nov 13 at 14:28
lubgr
10.2k21745
Don't use c function pointers, butstd::functioninstead
– hellow
Nov 13 at 14:29
13
@hellow no.std::functionis a heavywieght type-erased container for callable objects. If a function pointer is enough, use that.
– Quentin
Nov 13 at 14:32
add a comment |
16
16
16
You can bind a pointer to both functions in two variables and swap those.
void (*f1)() = printA;
void (*f2)() = printB;
f1(); // output: a
f2(); // output: b
std::swap(f1, f2);
f1(); // output: b
f2(); // output: a
answered Nov 13 at 14:28
lubgr
10.2k21745
You can bind a pointer to both functions in two variables and swap those.
void (*f1)() = printA;
void (*f2)() = printB;
f1(); // output: a
f2(); // output: b
std::swap(f1, f2);
f1(); // output: b
f2(); // output: a
answered Nov 13 at 14:28
lubgr
10.2k21745
answered Nov 13 at 14:28
lubgr
10.2k21745
answered Nov 13 at 14:28
lubgr
10.2k21745
answered Nov 13 at 14:28
lubgr
10.2k21745
10.2k21745
Don't use c function pointers, butstd::functioninstead
– hellow
Nov 13 at 14:29
13
@hellow no.std::functionis a heavywieght type-erased container for callable objects. If a function pointer is enough, use that.
– Quentin
Nov 13 at 14:32
add a comment |
Don't use c function pointers, butstd::functioninstead
– hellow
Nov 13 at 14:29
13
@hellow no.std::functionis a heavywieght type-erased container for callable objects. If a function pointer is enough, use that.
– Quentin
Nov 13 at 14:32
Don't use c function pointers, but
std::function instead– hellow
Nov 13 at 14:29
Don't use c function pointers, but
std::function instead– hellow
Nov 13 at 14:29
13
13
@hellow no.
std::function is a heavywieght type-erased container for callable objects. If a function pointer is enough, use that.– Quentin
Nov 13 at 14:32
@hellow no.
std::function is a heavywieght type-erased container for callable objects. If a function pointer is enough, use that.– Quentin
Nov 13 at 14:32
add a comment |
7
You need to wrap them in objects (or pointers to functions):
std::function<void()> myprintA = printA;
std::function<void()> myprintB = printB;
std::swap(myprintA, myprintB);
myprintA();
myprintB();
Otherwise, you are working with symbols themselves, and you can't swap this.
answered Nov 13 at 14:28
Matthieu Brucher
11.8k22138
add a comment |
7
You need to wrap them in objects (or pointers to functions):
std::function<void()> myprintA = printA;
std::function<void()> myprintB = printB;
std::swap(myprintA, myprintB);
myprintA();
myprintB();
Otherwise, you are working with symbols themselves, and you can't swap this.
answered Nov 13 at 14:28
Matthieu Brucher
11.8k22138
add a comment |
7
7
7
You need to wrap them in objects (or pointers to functions):
std::function<void()> myprintA = printA;
std::function<void()> myprintB = printB;
std::swap(myprintA, myprintB);
myprintA();
myprintB();
Otherwise, you are working with symbols themselves, and you can't swap this.
answered Nov 13 at 14:28
Matthieu Brucher
11.8k22138
You need to wrap them in objects (or pointers to functions):
std::function<void()> myprintA = printA;
std::function<void()> myprintB = printB;
std::swap(myprintA, myprintB);
myprintA();
myprintB();
Otherwise, you are working with symbols themselves, and you can't swap this.
answered Nov 13 at 14:28
Matthieu Brucher
11.8k22138
answered Nov 13 at 14:28
Matthieu Brucher
11.8k22138
answered Nov 13 at 14:28
Matthieu Brucher
11.8k22138
answered Nov 13 at 14:28
Matthieu Brucher
11.8k22138
11.8k22138
add a comment |
add a comment |
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5
This seems to be a possible instance of the XY problem. What is the root goal you're trying to achieve by swapping
ExitProcess?– Andrey Akhmetov
Nov 13 at 14:26
@AndreyAkhmetov I am participating in some kind of competition in which everyone want to make the enemy process stop and there is the option of using DLL injection and using the ExitProcess function in order to make my process exit so I want to avoid the option of others to use the function in my process.
– Mor Ben-Yosef
Nov 13 at 14:31