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警告:派任使点员在无放电的情况下感到愤怒
原标题:Warning: assignment makes integer from pointer without a cast
  • 时间:2011-09-12 07:56:26
  •  标签:
  • c

我正试图分两阵列中的内容,但在分配其价值使之产生这一错误时。

“警告:派任使点人陷入无 cast状态”。

为什么我去了,我如何解决这个问题?

for(i=1;i<3;++i)
    {
        for(j=3-1;j>=1;--j)
           {
                if(a[j-1]>a[j])
                   {
                         k = a[j-1];
                          a[j-1]=a[j];
                         a[j] = k;
                   }
           }
     }
问题回答

你们在获取这一要素方面失去了一个指数。 a[j], a[j-1]> bea[i][j], a[i][j-1] in all place. a[i] 是2D阵列中一行的第一个要素的指点。 <代码>a[i][j]为索引i, j

注:虽然问题没有具体指明,但我几乎确信,<代码>i, j, k 是星体,<代码>a是2D integer阵列。

警告是,在一份派任声明中,从点到点火的不安全转换。 这是你法典中的任何发言。 你们应当提供变式的类型申报,以便发现警报来自何处。





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