函数题第三题
void sum_diff( float op1, float op2, float *psum, float *pdiff )
{ *psum=op1+op2;
pdiff=op1-op2; }
psum,pdiff是地址psum,pdiff是值,所以对op1,op2做和差分别交给psum,*pdiff。*psum,*pdiff分别指向op1,op2的和差.