C++ 如何将程序执行跳转到C中的特定地址？
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How to jump the program execution to a specific address in C?
提问by Sam
I want the program to jump to a specific address in memory and continue execution from that address. I thought about using gotobut I don't have a label rather just an address in memory.
我希望程序跳转到内存中的特定地址并从该地址继续执行。我想过使用，goto但我没有标签，而只是内存中的地址。
There is no need to worry about return back from the jump address.
不用担心从跳转地址返回。
edit: using GCC compiler
编辑：使用 GCC 编译器
回答by Randall Cook
Inline assembly might be the easiest and most "elegant" solution, although doing this is highly unusual, unless you are writing a debugger or some specialized introspective system.
内联汇编可能是最简单和最“优雅”的解决方案，尽管这样做非常不寻常，除非您正在编写调试器或一些专门的内省系统。
Another option might be to declare a pointer to a void function (void (*foo)(void)), then set the pointer to contain your address, and then invoke it:
另一种选择可能是声明一个指向 void 函数 ( void (*foo)(void)) 的指针，然后将该指针设置为包含您的地址，然后调用它：
void (*foo)(void) = (void (*)())0x12345678;
foo();
There will be things pushed on the stack since the compiler thinks you are doing a subroutine call, but since you don't care about returning, this might work.
由于编译器认为您正在执行子例程调用，因此将有一些东西压入堆栈，但由于您不关心返回，这可能会起作用。
回答by Chris Dodd
gcc has an extension that allows jumping to an arbitrary address:
gcc 有一个允许跳转到任意地址的扩展：
void *ptr = (void *)0x1234567; // a random memory address
goto *ptr; // jump there -- probably crash
This is pretty much the same as using a function pointer that you set to a fixed value, but it will actually use a jump instruction rather than a call instruction (so the stack won't be modified)
这与使用您设置为固定值的函数指针几乎相同，但它实际上将使用跳转指令而不是调用指令（因此堆栈不会被修改）
回答by Rob?
#include <stdio.h>
#include <stdlib.h>
void go(unsigned int addr) {
(&addr)[-1] = addr;
}
int sub() {
static int i;
if(i++ < 10) printf("Hello %d\n", i);
else exit(0);
go((unsigned int)sub);
}
int main() {
sub();
}
Of course, this invokes undefined behavior, is platform-dependent, assumes that code addresses are the same size as int, etc, etc.
当然，这会调用未定义的行为，是平台相关的，假设代码地址与 等的大小相同int。
回答by Andro
It should look something like this:
它应该是这样的：
unsigned long address=0x80;
void (*func_ptr)(void) = (void (*)(void))address;
func_ptr();
However, it is not a very safe operation, jumping to some unknown address will probably result in a crash!
但是，这不是一个很安全的操作，跳转到某个未知地址可能会导致崩溃！
回答by U007D
Since the question has a C++ tag, here's an example of a C++call to a function with a signature like main()--int main(int argc, char* argv[]):
由于这个问题有一个 C++ 标签，下面是一个C++调用具有像 main()-- 签名的函数的示例int main(int argc, char* argv[])：
int main(int argc, char* argv[])
{
auto funcAddr = 0x12345678; //or use &main...
auto result = reinterpret_cast<int (*)(int, char**)>(funcAddr)(argc, argv);
}
回答by ?lkerK
This is what I am using for my bootstrap loader(MSP430AFE253,Compiler = gcc,CodeCompeserStudio);
这就是我用于引导加载程序的内容（MSP430AFE253,Compiler = gcc,CodeCompeserStudio）；
#define API_RESET_VECT 0xFBFE
#define JUMP_TO_APP() {((void (*)()) (*(uint16_t*)API_RESET_VECT)) ();}
回答by Bahman
I Propos this code:
我建议这个代码：
asm(
"LDR R0,=0x0a0000\n\t" /* Or 0x0a0000 for the base Addr. */
"LDR R0, [R0, #4]\n\t" /* Vector+4 for PC */
"BX R0"
);
回答by Taylor Price
Do you have control of the code at the address that you intend to jump to? Is this C or C++?
您是否可以控制要跳转到的地址处的代码？这是 C 还是 C++？
I hesitantly suggest setjmp() / longjmp()if you're using C and can run setjmp()where you need to jump back to. That being said, you've got to be VERY careful with these.
我犹豫地建议setjmp() / longjmp()你是否使用 C 并且可以运行setjmp()你需要跳回的地方。话虽如此，你必须非常小心这些。
As for C++, see the following discussion about longjmp()shortcutting exception handling and destructors destructors. This would make me even more hesitant to suggest it's use in C++.
至于 C++，请参阅以下有关longjmp()快捷方式异常处理和析构函数析构函数的讨论。这会让我更加犹豫是否建议在 C++ 中使用它。
