Question

练习它获取一个字符串数组，并使用指针将它们连接在一个新字符串中。我通过对每个字符串数组索引处的值进行硬编码来使其工作，但是如果数组没有元素，我无法弄清楚如何使它工作。我想我需要完全重构我的代码...

#include <stdio.h>
#include <stdlib.h>

int main(int argc, const char * argv[]) {
    char *newArray[3];

    //original code didn't have newArray as a pointer, and had the following:

    //char *firstString = &newArray[0];             This was pointing to the value at the address of newArray[0] which was null at this point in time

    //firstString = "Knicks";                       This was reassigning the value of firstString to "Knicks" instead of pointing to the value located at the proper index address.

    //this code correctly assigns values to each index of the array (which itself is just a pointer that is pointing to these values

    newArray[0] = "Knicks";
    newArray[1] = "Warriors";
    newArray[2] = "Bulls";

    //this code correctly assigns pointers that point to the index values of the array, which are all stated above.
    char *firstString = newArray[0];
    char *secondString = newArray[1];
    char *thirdString = newArray[2];



    int firstStringCount = 0;
    int secondStringCount = 0;
    int thirdStringCount = 0;

    //count length of first element
    for (int i = 0; i < 1000; i++) {
        if (firstString[i] == '\0') {
            break;
        }
        else {
            firstStringCount++;
        }
    }

    //count length of second element
    for (int i = 0; i < 1000; i++) {
        if (secondString[i] == '\0') {
            break;
        }
        else {
            secondStringCount++;
        }
    }

    //count length of third element
    for (int i = 0; i < 1000; i++) {
        if (thirdString[i] == '\0') {
            break;
        }
        else {
            thirdStringCount++;
        }
    }

    //int len = firstStringCount + secondStringCount + thirdStringCount;

    char *concatenatedString = malloc(firstStringCount + secondStringCount + thirdStringCount);

    for (int i = 0; i < firstStringCount; i++) {
        concatenatedString[i] = firstString[i];
    }

    for (int i = 0; i < secondStringCount; i++) {
        concatenatedString[i+firstStringCount] = secondString[i];
    }

    for (int i = 0; i < thirdStringCount; i++) {
        concatenatedString[i+firstStringCount+secondStringCount] = thirdString[i];
    }

    //add in null value
    concatenatedString[firstStringCount + secondStringCount + thirdStringCount] = '\0';


    printf("%s\n", concatenatedString);
    printf("%i\n", firstStringCount);
    printf("%i\n", secondStringCount);
    printf("%i\n", thirdStringCount);
    return 0;
}

Answer 1

当您有空字符串时，您的程序将起作用。这与没有数组元素不同，因为每个数组元素中仍然必须有一个\0终结符，例如将使用

创建

newArray[0] = "";

但是你没有为目标的字符串终止符分配足够的内存。使用1 +

char *concatenatedString = malloc(1 + firstStringCount + secondStringCount +
                                                         thirdStringCount);

你现在有了

终结者的空间

concatenatedString[firstStringCount + secondStringCount + thirdStringCount] = '\0';

使基于数组的程序在没有元素的情况下工作

1 个答案: