不是我错就是微软错，VC6.0中FLOAT 型二维数组问题。

[WJZMLSH]

UpdateData(TRUE);

static float qicha[21][5] = {{0.0f,0.0f,0.0f,0.0f,0.0f},{0.0f,0.0f,0.1f,0.1f,0.1f},{0.0f,0.1f,0.1f,0.1f,0.2f},{0.1f,0.1f,0.2f,0.2f,0.3f},
{0.1f,0.1f,0.2f,0.3f,0.3f},{0.1f,0.2f,0.3f,0.3f,0.4f},{0.1f,0.2f,0.3f,0.4f,0.5f},{0.1f,0.2f,0.4f,0.5f,0.6f},{0.1f,0.3f,0.4f,0.5f,0.7f},{0.2f,0.3f,0.5f,0.6f,0.8f},
{0.2f,0.3f,0.5f,0.7f,0.8f},{0.2f,0.4f,0.6f,0.7f,0.9f},{0.2f,0.4f,0.6f,0.8f,1.0f},{0.2f,0.4f,0.7f,0.9f,1.1f},{0.2f,0.5f,0.7f,0.9f,1.2f},{0.2f,0.5f,0.7f,0.9f,1.2f},{0.3f,0.5f,0.8f,1.0f,1.3f},{0.3f,0.5f,0.8f,1.1f,1.3f},{0.3,0.6f,0.9f,1.1f,1.4f},{0.3f,0.6f,0.9f,1.2f,1.5f},{0.3f,0.6f,1.0f,1.3f,1.6f}
};
float (*p)[5];
p=qicha;

m_f9=m_f1-m_f10;
m_f3=m_f2-m_f11;






float m=m_f3-m_f9;
if(abs(m)>=2.0) return;
int mm=abs(m*10);
if(m<0){m_f8=m_f9-*(*(p+mm)+0);
m_f7=m_f9-*(*(p+mm)+1);
m_f6=m_f9-*(*(p+mm)+2);
m_f5=m_f9-*(*(p+mm)+3);
m_f4=m_f9-*(*(p+mm)+4);}
if(m>=0){m_f8=m_f9+*(*(p+mm)+0);
m_f7=m_f9+*(*(p+mm)+1);
m_f6=m_f9+*(*(p+mm)+2);
m_f5=m_f9+*(*(p+mm)+3);
m_f4=m_f9+*(*(p+mm)+4);}



UpdateData(FALSE);
其中MF1-MF11是对话框上的变量，MF1 MF2 MF10 MF11是要输入的数据.
求器差分配的，将数组中数按要求分配给MF3-MF9，但有个问题，若MF9或MF3其中有一个是0.4的话，且M是0.2时就会不是
你所求的，但我用同样的算法用在整型数据上就没问题，FLOAT换成DOUBLE也不行。
我用VC6.0英文专业与企业版都试过。
我只好改了算法实现。
感兴趣的我可以给源码看看，

[vcdream]

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thx