指针 (char*) 与 integer 错误比较

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聪明一点， and在最后一个国家的句子末尾加上和。

例如3个国家：

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 integer 错误。

char*是什么意思？

如何访问数组中的最后一个元素？

Answer 1

country countries[x]是一个string （在CS50中是char*的 typedef）， number_of_p是一个int ，你不能比较它们，它们是不同的类型，你可能想比较索引x ，一个可能的（和快速）修复您的代码，包括标点符号可能如下所示：

现场演示

#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

Output：

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 。

也就是说，类型说明符string是char *类型的别名，而这个数组声明

string countries[number_of_p];

是相同的

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 output 是

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

Answer 3

您可以使用一个 printf 执行此类操作，使用预定义的字符串作为分隔符。 使用单个 printf 而不是特殊套管，列表的尾部通常更易于维护。 许多人对三元运算符感到困惑。 这是两种不同方式的方法，首先使用开关，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" : ", ");
        }
}