使用指针扫描指针数组

This is an example to how check the dimension of an array of char pointers:

这是如何检查char指针数组的维度的示例:

#include <stdio.h>
int main(void){

    // prototype
    int checkDim( char **array);

    char *days[] = {      //array of pointers
        "Sunday",         //days[0] = {'S','u','n','d','a','y','\0'}
        "Monday",         //days[1] = {....}
        "Tuesday",        //days[2] = {....}
        "Wednesday",      //days[3] = {....}
        "Thursday",       //days[4] = {....}
        "Friday",         //days[5] = {....}
        "Saturday"        //days[6] = {....}
    };

    int dim;
    dim = checkDim( days );

    printf("Number of days = %i \n", dim );

    return 0;
}

int checkDim(  char **array ){

    char **ptr = array;

    while(*ptr){
        ++ptr;
    }

    return ptr - array;
}

Respectively:

days[0] is a pointer to char (another array),
days[1] is a pointer to char (another array)
...
days[i] is a pointer to char (another array)

days [0]是指向char(另一个数组)的指针,days [1]是指向char的指针(另一个数组)... days [i]是指向char的指针(另一个数组)

With a while I can scan the array containing the pointers:

有一段时间我可以扫描包含指针的数组:

char **ptr = array; //

while(*ptr){
    ++ptr;
}

How I can can stop on the first element days[0] and scan every letter to check the length of days[0] (subsequently days[1], days[2] etc...)?

I would like a printf like this (for example):

我想像这样的printf(例如):

> Number of days 7.

>天数7。

> Sunday is long 6 letters

>星期天长6封信

> Monday is long 6 letters

>周一长6封信

> Tuesday is long 7 letters

>周二长7封信

etc..

1 个解决方案

#1

1st of all:

首先:

checkDim() as it stands invokes undefined behaviour by accessing the pointer array days out of bounds when looping past the pointer for "Saturday".

checkDim()因为它在循环超过“星期六”的指针时通过访问指针数组的日期来调用未定义的行为。

It's just bad luck the code seems to run.

代码似乎运行起来运气不好。

To fix this do explicitly introduce a NULL value at the end of days to functions as a sentinel:

要解决此问题,请在日期结束时显式引入NULL值作为标记:

  char * days[] = {      //array of pointers
    "Sunday",         //days[0] = {'S','u','n','d','a','y','\0'}
    ...
    "Saturday",        //days[6] = {....}
    NULL
  };

---

To answer your question:

回答你的问题:

To print the length of each "string" during iteration you code do this:

要在迭代期间打印每个“字符串”的长度,您可以执行以下操作:

  while (*ptr)
  {
    printf("'%s' has length %zu.\n", *ptr, strlen(*ptr));
    ++ptr;
  }

Please note that strlen() returns size_t (not int) to using the conversion specifiers d or u is wrong here. If zu is not defined by the compiler in use (VC) you want to use lu instead.

请注意,strlen()返回size_t(不是int)以使用转换说明符d或u在这里是错误的。如果zu未被使用中的编译器(VC)定义,则需要使用lu。

---

Two more things:

还有两件事:

1. Placement of prototypes

   原型的放置

   Prototypes should be at global scope.

   原型应该在全球范围内。

   So change this:

   所以改变这个:

   int main(void)
   {
     // prototype
     int checkDim( char **array);
   

   to be:

   // prototype
   int checkDim( char ** array);
   
   int main(void)
   {
   

2. Qualify argument staying constant as const

   限定参数保持常量为常量

   checkDim does not change days, so help the compiler to create the best code by telling this to it.

   checkDim不会改变天数,因此请通过告诉它来帮助编译器创建最佳代码。

   Change this:

   int checkDim(char **array) 
   

   to be:

   int checkDim(const char ** array)