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从这个问题我了解到，您确实不应该导出局部变量的地址并在声明它的函数之外使用它。

但是，在我看来，K&R 在下面显示的程序中违反了这条规则，该程序取自他们的书，p。108.

我正在查看lineptr[nlines++] = p;函数内部的行readlines。为什么可以在这里“导出”p并稍后在外面使用readlines？

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

#define MAXLINES 5000

char *lineptr[MAXLINES];

int readlines(char *lineptr[], int nlines);
void writelines(char *lineptr[], int nlines);

void qsort(char *lineptr[], int left, int right);

int main(int argc, char *argv[])
{
     int nlines;

     if((nlines = readlines(lineptr, MAXLINES)) >= 0) {
          qsort(lineptr, 0, nlines-1);
          writelines(lineptr, nlines);
          return 0;
     } 
     else {
          printf("error: input too big to sort\n");
          return 1;
     }
}


#define MAXLEN 1000
int getline(char *, int);
char *alloc(int);

int readlines(char *lineptr[], int maxlines)
{
    int len, nlines;
    char *p, line[MAXLEN];

    nlines = 0;
    while((len = getline(line, MAXLEN)) > 0)
       if(nlines >= maxlines || (p = alloc(len)) == NULL)
          return -1;
       else {
            line[len-1] = '\0';
            strcpy(p, line);
            lineptr[nlines++] = p;
       }
    return nlines;
}

void writelines(char *lineptr[], int nlines)
{
     while(nlines -- > 0)
         printf("%s\n", *lineptr++);
}

int getline(char s[], int lim)
{
  int c, i;

  for (i = 0; i < lim - 1 && (c = getchar()) != EOF && c != '\n'; i++)
    s[i] = c;                                                         
  if (c == '\n') {
    s[i++] = c;   
  }
  s[i] = '\0';
  return i;
}

#define ALLOCSIZE 10000

static char allocbuf[ALLOCSIZE];
static char *allocp = allocbuf;

char *alloc(int n)
{
     if(allocbuf + ALLOCSIZE - allocp >= n) {
          allocp +=n;
          return allocp - n;
     }
     else 
          return 0;
}

void swap(char *v[], int i, int j)
{
     char *temp;

     temp = v[i];
     v[i] = v[j];
     v[j] = temp;
}

void qsort(char *v[], int left, int right) {

    int i, last;

    if(left >= right) 
       return;

    swap(v, left, (left+right)/2);
    last = left;

    for(i = left + 1; i <= right; i++)
      if(strcmp(v[i], v[left]) < 0)
         swap(v, ++last, i);

    swap(v, left, last);
    qsort(v, left, last-1);
    qsort(v, last+1, right);
}
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1 回答 1

6

在：

lineptr[nlines++] = p;

the value of p is stored, not its address. The address is &p. Of course, the value of p happens to be an address, since p is a pointer, and the value of a pointer represents an address. But that has no consequence here. The rule is still being followed; no address of any local variable has been stored anywhere outside the function, and the value of p is not the address of a local variable.

If you follow the call chain you can determine that the value of p can be either 0, or be an address somewhere inside allocbuf. And allocbuf is not a local variable. It's a static variable at file scope.

于 2013-06-18T07:27:51.767 回答