如何在不使用其他数组的情况下在C的指针的字符串数组中交换两个字符串? [英] how can I swap two strings in a string-array of pointers in C without using other arrays?

问题描述

int main(){

char* ants[5]={"red ant" ,"blue ant" ,"green ant" ,"yellow ant" ,"white ant"};

.... }

如果要交换红蚂蚁" 和白蚁" ，如何交换两个字符串?有人可以帮我写代码(根本不使用另一个数组)吗?

how can I swap two strings if I want to swap "red ant" and "white ant" ? can someone help me write the code (without using another array at all)???

推荐答案

用于蚂蚁类型的字符串是 String-Literals ，除不符合标准的编译器外，其他所有编译器都是只读的(非-可变).字符串文字的类型为字符数组，其长度设置为包含字符和'\ 0'(以零结尾的字符).字符串文字的标准部分为 C11标准-6.4.5字符串文字

The strings used for the types of ants are String-Literals and on all but non-conforming compilers are read-only (non-mutable). A string literal has the type array of characters with the length set to contain the characters and the '\0' (nul-terminating characer). The standard section for string literals is C11 Standard - 6.4.5 String literals

您不能更改字符串文字的内容，但是可以更改哪个指针指向哪个字符串字面量.就您而言，您有:

You cannot change the contents of the string literals, but you can change which pointer points to which string-literal. In your case you have:

    char *ants[5]={"red ant" ,"blue ant" ,"green ant" ,"yellow ant" ,"white ant"};

类型为指向的指针数组 char .因此，您有一个字符指针数组.数组中的每个指针都保存 ants 中字符串之一的地址.要更改 ant 中字符串的顺序，您必须交换数组中指针所持有的地址.

Which has the type Array of Pointers to char. So you have an array of character pointers. Each of the pointers in the array holds the address of one of the strings in ants. To change the order of the strings in ants you have to swap the addresses held by the pointers in your array.

交换红色和蓝色蚂蚁的简单例子可能是:

A trivial example to swap the red and blue ants could be:

#include <stdio.h>

int main(){

    char *ants[5]={"red ant" ,"blue ant" ,"green ant" ,"yellow ant" ,"white ant"};
    int n = sizeof ants / sizeof *ants;         /* number of ants */
    
    char *t = ants[0];                          /* temp pointer holding 1st address */
    ants[0] = ants[1];                          /* assign 2nd address to 1st pointer */
    ants[1] = t;                                /* assign temp pointer to 2nd */
    
    for (int i = 0; i < n; i++)                 /* output swapped ants */
        puts (ants[i]);
}

使用/输出示例

$ ./bin/swap_ants
blue ant
red ant
green ant
yellow ant
white ant

如果您想编写 swap()函数以交换指针，则不能仅将指针作为参数传递，例如，您不能编写:

If you wanted to write a swap() function to swap the pointers, you cannot just pass the pointers as arguments, e.g., you could NOT write:

swap (char *a, char *b)

，然后在函数中交换 a 和 b .为什么?指针 a 和 b 是实际指针的副本.它们是 swap()函数的局部变量.因此，在 main()中将看不到您在函数中对其执行的任何操作.

and then swap a and b in the function. Why? The pointers a and b are copies of the actual pointers. They are local variables for the swap() function. Therefore, nothing you do to them in the function would be seen back in main().

您可以提供指针的地址作为参数，然后在这两个地址处交换指针，并且更改将在 main()中被看到.通过传递原始指针的实际地址，可以对原始地址处的值进行任何更改，并且可以在 main()中使用.因此，用于两个指针的 swap()函数可以写为:

You CAN provide the address of the pointers as the arguments and then swap the pointers at those two addresses and the changes WILL be seen back in main(). By passing the actual address of the original pointer, any changes made are made to the values at the original address and are available back in main(). So a swap() function for two pointers could be written as:

/** swap pointers, changing pointer at address */
void swapptr (char **a, char **b)
{
    char *t = *a;           /* pointer at a's address saved in t */
         *a = *b;           /* assign pointer from b to a */
         *b = t;            /* assign temp pointer t to b */
}

现在，如果您想交换红色和蓝色的蚂蚁，您可以简单地调用 swapptr()函数，传递每个指针的地址，例如

Now if you wanted to swap the red and blue ants, you could simply call the swapptr() function passing the address of each of the pointers, e.g.

#include <stdio.h>

/** swap pointers, changing pointer at address */
void swapptr (char **a, char **b)
{
    char *t = *a;           /* pointer at a's address saved in t */
         *a = *b;           /* assign pointer from b to a */
         *b = t;            /* assign temp pointer t to b */
}

int main(){

    char *ants[5]={"red ant" ,"blue ant" ,"green ant" ,"yellow ant" ,"white ant"};
    int n = sizeof ants / sizeof *ants;         /* number of ants */
    
    swapptr (ants, ants + 1);
    
    for (int i = 0; i < n; i++)                 /* output swapped ants */
        puts (ants[i]);
}

(相同的输出)

注意:您必须注意类型. ants (即 ants + 0 )和 ants + 1 是指向字符串文字的指针，它们是指向chars数组的指针.但是在访问时，C语言中的Type数组将转换为指向第一个元素的指针. C11标准-6.3.2.1其他操作数-左值，数组和函数标识符(p3)因此，每个都是指向每个字符中第一个字符的指针.因此，每个元素的开头都是 char ** 类型.所以这就是为什么您可以传递 ants + 0，ants + 1 来交换第一和第二(红色和蓝色)蚂蚁的原因.

Note: you have to pay attention to the types. ants (which is ants+0), and ants + 1 are pointers to your string literals, they are pointers to an array of chars. But on access, an array of Type in C is converted to a pointer to the first element. C11 Standard - 6.3.2.1 Other Operands - Lvalues, arrays, and function designators(p3) So each and is a pointer to a pointer to the first character in each. So each element has the type char** to begin with. So that's why you can pass ants + 0, ants + 1 to swap the first and second (red and blue) ants.

仔细检查一下，如果还有其他问题，请告诉我.

Look things over and let me know if you have further questions.

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