C:指针和数组

Question

我的一段代码片段。我下面的功能接受来自用户的文本。反转整个输入，然后尝试将反转字符串的 ASCII 表示形式存储在名为 ascii_array 的数组中。然而，就目前的代码而言，它只接受反转字符串的第一个字符，递增并覆盖 ascii_array 的第一个位置。我尝试了几种方法来增加指针以将字符存储在 ascii_array 中，但出现错误。该数组应包含 ASCII 数组中反转字符串的 ASCII 表示。

例子:输入:你好!

反转字符串的 Ascii 表示:!elloh

预期输出:33 111 108 108 101 104

我得到的输出:33

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

void encrypt(char inputText[], int inputLength);

int main(void)
{
    char word[512];
    int length;
    printf("Enter word: ");
    fgets(word,512,stdin);
    length = strlen(word);
    encrypt(word,length);
    return 0;

}

void encrypt(char inputText[], int inputLength)
{

    printf("%s\n",inputText);
    char *begin = inputText;
    char *end = inputText + inputLength - 1;
    char temp;
    while(end > begin)
    {
        temp = *end;
        *end-- = *begin;
        *begin++ = temp;

    }

    char *ascii_pointer;
    char temporary;
    ascii_pointer = inputText + 1;
    temporary = *ascii_pointer;

    int *p;
    int array[512];
    p = array;
    while(ascii_pointer < inputText + inputLength)
    {
        printf("INPUTTEXT: ascii_pointer: %s\n " , ascii_pointer);
        temporary = *ascii_pointer++;
        printf("temporary: %c\n " , temporary);
        *p = temporary++;
        // I think my logic is incorrect fron this point.
        //*p++ = temporary++;
        //*p++;
        printf("asci_pointer: %c\n " ,*p);
        printf("ascii_array: %d\n ", *array);
        printf("\n");
    }

    printf("\n");
    printf("END: %p:%c\n", p, *p);
    printf("END--- MAYBE GARBAGE VALUE: place holder: %c\n ",temporary);
    printf("END: ascii_array: %d\n ", *array);
    return;

}

Answer 1

I want to have an array of integers. Because later on in the code I will be adding values to those integers.

如果您想将 char * 转换为整数数组，没有太多工作要做。 char 已经是一个 1 字节的整数。您可以简单地保持原样并对其进行数学计算。

例如……

int main(void) {
    char string[] = "Hello";

    // Iterate through the string using the pointer until we see null.
    // It avoids having to use strlen() which scans the whole string.
    for( char *tmp = string; tmp[0] != '\0'; tmp++ ) {
        // Print the char as an integer.
        printf("%d\n", tmp[0]);
    }
}

$ ./test
72
101
108
108
111

如果您想操纵这些数字，请继续!这是一个将每个字符加 2 的示例。

int main(void) {
    char string[] = "Hello";

    // Iterate through the string using the pointer until we see null.
    // It avoids having to use strlen() which scans the whole string.
    for( char *tmp = string; tmp[0] != '\0'; tmp++ ) {
        tmp[0] += 2;
    }

    puts(string);
}

$ ./test
Jgnnq

请记住，char 是一个字节，最多只能存储 127。

所以您的加密函数需要做的就是反转字符串。

void encrypt(char inputText[], int inputLength)
{
    printf("%s\n",inputText);
    char *begin = inputText;
    char *end = inputText + inputLength - 1;
    char temp;
    while(end > begin)
    {
        temp = *end;
        *end-- = *begin;
        *begin++ = temp;

    }
}

---

如果您真的想要将字符串转换为整数数组，一个简单的循环就可以了。

int main(void) {
    char string[] = "Hello";

    // Allocate an integer array to hold each integer in the string.
    // This ignores the null byte.
    int *copy = malloc( strlen(string) * sizeof(int) );

    size_t len = strlen(string);

    // Simple iteration to copy the string. A char will always
    // be smaller than an int so the assignment just works.
    for( size_t i = 0; i < len; i++ ) {
        copy[i] = string[i];
    }

    // Print out each integer in the new array.
    for( size_t i = 0; i < len; i++ ) {
        printf("%d\n", copy[i]);
    }

    free(copy);
}