指针（char*）与整数错误的比较

Question

所以基本上用户在一个变量中输入他们访问过多少个国家，然后我把它作为数组的大小。

之后，我使用for 循环列出所有访问过的国家/地区。但我想让我的代码更智能一点，并将and 放在最后一个国家/地区的句子末尾。

例如3国家：

You visited Japan, Korea and Canada.
                 ^       ^^^

#include <stdio.h>
#include <cs50.h>


int main(void)
{
    int number_of_p = get_int("How many countries did you visit?\n");
    string countries[number_of_p];

    for (int x = 0; x < number_of_p; x++)
    {
        countries[x] = get_string("What countries did you visit?\n");
    }
    printf("You visited %i countries, including: ", number_of_p);
    for (int x = 0; x < number_of_p; x++)
    {

        printf("%s, ", countries[x]);
        /* looks for last element in arrays, inserts 'and' 
        to make it look grammatically correct. */

        if (countries[x] == number_of_p - 1 ) // ERROR HERE
        {
            printf(" and ");
        }

    }
    printf(".\n");

}

我正在比较指针 (char*) 和整数错误。

char* 是什么意思？

如何访问数组中的最后一个元素？

Answer 1

countries[x]是string（在CS50中是char*的typedef），number_of_p是int，你不能比较它们，它们是不同的类型，你可能有想要比较索引x，这是一种可能（且快速）的代码修复方法，包括标点符号如下所示：

Live demo

#include <stdio.h>
#include <cs50.h>

int main(void)
{
    int number_of_p = get_int("How many countries did you visit?\n");
    string countries[number_of_p];

    for (int x = 0; x < number_of_p; x++)
    {
        countries[x] = get_string("What countries did you visit?\n");
    }
    printf("You visited %i countries, including: ", number_of_p);
    for (int x = 0; x < number_of_p; x++)
    {

        printf("%s", countries[x]);
        if(x < number_of_p - 2){
            printf(", ");
        }
   
        if (x == number_of_p - 2)
        {
            printf(" and ");
        }  
    }
    printf(".\n");
}

输入：

3
Japan
Korea
Canada

输出：

You visited 3 countries, including: Japan, Korea and Canada.

Answer 2

if 语句中的条件

    if (countries[x] == number_of_p - 1 )

没有意义。左操作数countries[x] 的类型为char *，而右操作数的类型为int。

即类型说明符string 是类型char * 的别名，而这个数组声明

string countries[number_of_p];

与

相同

char * countries[number_of_p];

所以你有一个指向字符串的指针数组。

C 中 char * 类型的别名 string 定义如下

typedef char * string;

循环看起来像

for (int x = 0; x < number_of_p; x++)
{
    if ( x != 0 )
    {
        putchar( ',' );
        putchar( ' ' );

        if ( x ==  number_of_p - 1 )
        {
            printf( "%s ", "and " );
        }
    }
    printf("%s", countries[x]);
}

这是一个演示程序

#include <stdio.h>

int main(void) 
{
    enum { number_of_p = 3 };
    char *countries[number_of_p] =
    {
        "Japan", "Korea", "Canada"
    };
    
    printf( "You visited %i countries, including: ", number_of_p );
    
    for (int x = 0; x < number_of_p; x++)
    {
        if ( x != 0 )
        {
            putchar( ',' );
            putchar( ' ' );

            if ( x ==  number_of_p - 1 )
            {
                printf( "%s ", "and" );
            }
        }
        printf("%s", countries[x]);
    }
    
    return 0;
}

它的输出是

You visited 3 countries, including: Japan, Korea, and Canada

Answer 3

你可以用一个 printf 来做这种事情，使用一个预定义的字符串作为分隔符。使用单个 printf 而不是特殊的大小写，列表的尾部通常更容易维护。许多人对三元运算符感到困惑。这里有两种不同的方法，首先使用 switch，然后使用三元运算符：

#include<stdio.h>

int main(void)
{
        char *countries[] = { "foo", "bar", "baz" };
        int number_of_p = sizeof countries / sizeof *countries;
        /* First loop, using a switch */
        for( int x = 0; x < number_of_p; x++ ) {
                char *div = ", ";
                switch( x ){
                case sizeof countries / sizeof *countries - 2: div = ", and "; break;
                case sizeof countries / sizeof *countries - 1: div = ".\n"; break;
                }
                printf("%s%s", countries[x], div);
        }
        /* Same loop as above, with if/else instead of the switch */
        for( int x = 0; x < number_of_p; x++ ) {
                char *div = ", ";
                if( x == number_of_p - 2 ) {
                        div = ", and ";
                } else if( x == number_of_p - 1 ) {
                        div = ".\n";
                }
                printf("%s%s", countries[x], div);
        }
        /* Same loop as above, written less verbosely */
        for( int x = 0; x < number_of_p; x++ ) {
                printf("%s%s", countries[x], x == number_of_p - 2 ? ", and " :
                        x == number_of_p - 1 ? ".\n" : ", ");
        }
}