复制数组使用原始数组修改函数的行为

Question

Edit: I've edited my code to add the 'extern"C"' part (not sure if that makes a difference)

So I have this small code:

extern "C"{
   void foo(int* nR,int* mR,float* x,float* out){       //&& 
       const int n=*nR,m=*mR;
       other_foo(n,m,x,out);
   }
}

that works fine. But now I want to copy the n-array of float x before passing it to the function other_foo (since other_foo will be altering x and I want to keep a copy).

if I do like so, all works fine:

extern "C"{
   void foo(int* nR,int* mR,float* x,float* out){       //&& 
       const int n=*nR,m=*mR;
       float y[n];
       for(int i=0;i<n;i++) y[i]=x[i];
       other_foo(n,m,x,out);
   }
}

but if I do like so:

extern "C"{
   void foo(int* nR,int* mR,float* x,float* out){       //&& 
       const int n=*nR,m=*mR;
       float y[n];
       std::copy(x,x+n,y);
       other_foo(n,m,x,out);
   }
}

all hell breaks lose: the output of other_foo is not the same anymore!

My question is, of course, why?

Answer 1

There is no difference between these two code snippets relative to the result of copying.

void foo(int* nR,int* mR,float* x,float* out){      //&& 
    const int n=*nR,m=*mR;
    float y[n];
    for(int i=0;i<n;i++)    y[i]=x[i];
    other_foo(n,m,x,out);
}


void foo(int* nR,int* mR,float* x,float* out){      //&& 
    const int n=*nR,m=*mR;
    float y[n];
    std::copy(x,x+n,y);
    other_foo(n,m,x,out);
}

The both copy n elements from x to y.

However this code is not C++ compliant. The size of an array shall be a constant expression known at compile time. So it would be more correctly to allocate the array dynamically.

For example

void foo(int* nR,int* mR,float* x,float* out){      //&& 
    int n = *nR, m = *mR;
    float *y = new float[n];
    std::copy( x, x+n, y );
    other_foo( n, m, x, out );
    // other code  maybe including delete [] y
}

I think that the problem is not in this function. It seems that the problem is in called function other_foo