指针下标从负值开始 [英] pointer subscript starts from negative value

问题描述

int nrow=5,ncol=7,i,j;
float **ptr;
/*allocation*/
ptr=(float **) malloc(nrow*sizeof(float*));
ptr[0]=(float *) malloc(nrow*ncol*sizeof(float));
/*initialize*/
for (i=0;i<nrow;i++)
    for (j=0;j<ncol;j++) ptr[i][j]=0.0;

在上述情况下，我们知道行从0开始并以nrow-1结尾，列从0开始并以ncol-1结尾.但是如何让行从-4开始并在nrow+3结束，又如何使列从-4开始并在ncol+3结束?

We know in the above case, the row starts from 0 and ends in nrow-1, the column starts from 0 and ends in ncol-1. But how can I let the row start from -4 and end in nrow+3, also let the column start from -4 and end in ncol+3?

补充代码:

#include <stdio.h>
#include <stdlib.h>

float *vector(int nl, int nh){
/* allocate a float vector with subscript range v[nl..nh] and initializing
       this vector, eg. vector[nl..nh]=0.0 */
    float *v;
    int i,NR_END=0;

    v=(float *)malloc((size_t) ((nh-nl+1+NR_END)*sizeof(float)));

    for (i=0;i<(nh-nl+1+NR_END);i++) v[i]=0.0;
    return v-nl+NR_END;
}

int main(int argc, char *argv[])
{
    int i,nrow=5, row1, row2;
    float *v;
    row1=-4;
    row2=nrow+3;
    v = vector(row1,row2);
    for (i=-4;i<(nrow+4);i++) {
        v[i]=(float)i; 
        printf("v[%d]=%f\n",i,v[i]);
    }
exit(0);
}

如果我运行上面的代码，它将得到正确的答案:

If I run the above code, it'll get the correct answer:

 v[-4]=-4.000000
 v[-3]=-3.000000
 v[-2]=-2.000000
 v[-1]=-1.000000
 v[0]=0.000000
 v[1]=1.000000
 v[2]=2.000000
 v[3]=3.000000
 v[4]=4.000000
 v[5]=5.000000
 v[6]=6.000000
 v[7]=7.000000
 v[8]=8.000000

推荐答案

比方说，您的数组有7个元素，其布局如下:

Let's say your array has 7 elements and is laid out as below:

+-----+-----+-----+-----+-----+-----+-----+
|     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+-----+-----+

如果指针指向数组的第一个元素，则可以将其索引为:

If a pointer points to the first element of the array, it can be indexed as:

p
|
v
+-----+-----+-----+-----+-----+-----+-----+
|     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+-----+-----+
p[0]  p[1]  p[2]  p[3]  p[4]  p[5]  p[6]  

如果指针指向数组中间的元素，则可以使用负值对其进行索引.

If a pointer points to an element in the middle of the array, it can be indexed using negative values.

                  p
                  |
                  v
+-----+-----+-----+-----+-----+-----+-----+
|     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+-----+-----+
p[-3] p[-2] p[-1] p[0]  p[1]  p[2]  p[3]  

如果指针指向最后一个元素之后的一个元素，则只能使用负值对其进行索引.

If the pointer points to one element past the last element, it can be indexed using only negative values.

                                          p
                                          |
                                          v
+-----+-----+-----+-----+-----+-----+-----+
|     |     |     |     |     |     |     |
+-----+-----+-----+-----+-----+-----+-----+
p[-7] p[-6] p[-5] p[-4] p[-3] p[-2] p[-1] 

先行者指向第一个元素之前的任何内容都是无效的.因此，无论指针指向元素的有效范围内的什么位置，有效索引都不能小于-7或大于6.

It is not valid for the pionter to point to anything before the first element. Hence, no matter where the pointer points to in the valid range of elements, a valid index cannot be less than -7 or greater than 6.

提出您的问题

但是如何让行从-4开始并以nrow+3结束，又如何使列从-4开始并以ncol+3结束?

But how can I let the row start from -4 and end in nrow+3, also let the column start from -4 and end in ncol+3?

您不能.如果指针指向数组的第5个元素，则可以使用-4作为有效索引，但是结束条件将为nrow-4/ncol-4. nrow + <some number>/ncol + <some number>永远不会是正确的结束索引.

You cannot. If a pointer points to the 5-th element of the array, you can use -4 as a valid index but then the end condition is going to be nrow-4/ncol-4. nrow + <some number>/ncol + <some number> will never be the correct end index.

float** ptr1 = &(5-th row of the array);
for ( int i = -4; i < nrow - 4; ++i )
{
   // OK to use ptr1[i];
   float* ptr2 = &(5-the element/column of the row)
   for ( int j = -4; j < ncol - 4; ++j )
   {
      // OK to use ptr2[j];
   }
}

这篇关于指针下标从负值开始的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！