c++ - 填充二维字符数组并访问每个元素

这是我在堆栈溢出中的第一个问题,所以如果有一些错误对此感到抱歉。我正在尝试填充 2d 字符数组,然后访问每个字母。我编译了我的代码,没有错误,但是当我尝试运行它时它不起作用。这是我的代码。

#include<iostream>
#include<string>
#include<stdlib.h>

using namespace std;


int main() {

    char ch[] = "Welcome text in a separate line.";
    char strWords[5][7];
    int counter = 0;
    int a = 0;

    for (int i = 0; i < sizeof(ch); i++) {

        if (ch[i] == ' ') {
            strWords[counter][a] = '\0';
            counter++;
            a = 0;
        }
        else
        {
            strWords[counter][a] += ch[i];
            a++;
        }

    }

    for (int i = 0; i <= 5; i++) {

        for (int a = 0; a <= 7; a++) {
            cout << strWords[i][a] << "  ";

        }
    }




    return 0;
}

回答1

您的代码有一些问题

int main() {

    char ch[] = "Welcome text in a separate line.";
    // char strWords[5][7]; <<<=== i would change to be larger that you need, just in case
    char strWords[20][20];
    int counter = 0;
    int a = 0;

    for (int i = 0; i < strlen(ch); i++) { // sizeof is wrong, you need strlen

        if (ch[i] == ' ') {
            strWords[counter][a] = '\0';
            counter++;
            a = 0;
        }
        else
        {
            //strWords[counter][a] += ch[i];
            strWords[counter][a] = ch[i]; // you do not need to try to concatenate, you are already walking down the buffer with 'a'
            a++;
        }

    }

    for (int i = 0; i < counter; i++) { // use 'counter' as it has the number of lines
       // since you 0 terminated the string you do not need to walk character by character

         cout << strWords[i] << "  ";

    }


    return 0;
}

您也没有检测和终止最后一个单词(因为它后面没有空格)。我会把它留给你。我显示的代码不打印单词“line”。

您确实应该进行测试以确保您不会超出单词的长度或数量。

另外,您最好使用 std::stringstd::vector

注意 - 如果为了实验,您确实想逐个字符地遍历以输出字符串,您应该查找终止的 '0' 字符并退出内部循环

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