旧代码：访问作为参数传递给另一个函数的函数的参数？
我必须维护一段旧代码，该代码在 64k 机器上编译时不再正常工作。
我有一个函数“solve”调用另一个函数“funct”作为指针传递：
intsolve(double*x,double xA,double xB,double zeps,double funct(double x,double*),... )
因此，“solve”可以与不同的可能函数一起使用，例如：
double isDgood(double D,double*Y);
在函数“solve”中，可以调用函数“funct”并使用以下方式访问其参数：
fA=funct(xA,(double*)(&funct+1));"
虽然我不熟悉这种语法，但我猜测开发人员假设未指定的参数只是被推入堆栈中。但是，该代码在 64k 平台上不再起作用。我应该如何更正该代码？
非常感谢您的帮助。
I have to maintain an old piece of code which does not work correctly anymore when compiled on 64k machines.
I have a function "solve" calling another function "funct" passed as a pointer:
int solve(double*x,double xA,double xB,double zeps,double funct(double x,double*),...)
Therefore, "solve" can be used with different possible functions as for example :
double isDgood(double D,double*Y);
From within the function "solve", it was possible to call the function "funct" and access its arguments using:
fA=funct(xA,(double*)(&funct+1));"
Although I am not familiar with this syntax, I guess that the developer was assuming that the unspecified arguments were just pushed in the stack. However, this code does not work anymore on 64k platforms. How can I correct this code? Should I specifically use Va_list?
Thanks you very much for your help.
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这是可怕的未定义行为。如果您想访问参数，则必须传递它们。
That's horrifically undefined behaviour. If you want to access the argument, you will have to pass them around.