c++ - 关于命令行参数/char* 的三个问题

Question

我的帖子分为三个部分:

1. My code
2. Example input and output
3. My three questions

我的代码:

#include <iostream>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <fstream>
#include <sstream>
#include <string>
#include <cstring>

using namespace std;

void deleteTrash(char*, char*);

const int kStr = 2;
const int kStrLen = 3;

int main(int argc, char* argv[])
{
    if (argc < 4) {
        cout << "Incorrect argument given." << endl;
        cout << "Try again." << endl;
        return 0;
    }

    cout << "PRINT argv[2]" << endl;
    cout << "-----" << endl;
    for (int i = 0; i < sizeof(argv[2]); i++) {
       cout << "Iterator: " << i << endl;
       cout << argv[2][i] << endl;
    }

    char* inputString;
    deleteTrash(argv[kStr], inputString);

    cout << "PRINT inputString" << endl;
    cout << "-----" << endl;
    for (int i = 0; i < sizeof(inputString); i++) {
        cout << i << endl;
        cout << inputString[i] << endl;
    }

    int strLen;
    stringstream num;
    num << argv[kStrLen];
    num >> strLen;

    if ( num.fail() ) {
        cout << "Incorrect argument given." << endl;
        cout << "Try again." << endl;
        return 0;
    }

    if ( strLen < sizeof(inputString) ) {
        cout << "Incorrect argument given." << endl;
        cout << "Try again." << endl;
        return 0;
    }

    return 0;
}

void deleteTrash(char* tempString, char* inputString) 
{
    int tempStringLen = sizeof(tempString);
    int newSize = 0;

    while (tempString[newSize] != '\0')
        newSize++;

    char newString[newSize + 1];

    int iterator = 0;

    while (tempString[iterator] != '\0') {
        newString[iterator] = tempString[iterator];
        iterator++;
    }

    newString[newSize] = '\0';

    cout << "PRINT newString" << endl;
    cout << "-----" << endl;
    for (int i = 0; i < sizeof(newString); i++) {
        cout << newString[i] << endl;
    }   

    inputString = newString;

    cout << "PRINT inputString" << endl;
    cout << "-----" << endl;
    for (int i = 0; i < sizeof(inputString); i++) {
        cout << "Iterator: " << i << endl;
        cout << inputString[i] << endl;
    }   

    return;
}

示例输入:

./hw1q5 4 W# 3

输出:

PRINT argv[2]
-----
Iterator: 0
W
Iterator: 1
#
Iterator: 2

Iterator: 3
3
Iterator: 4

Iterator: 5
T
Iterator: 6
E
Iterator: 7
R
PRINT newString
-----
W
#

PRINT inputString
-----
Iterator: 0
W
Iterator: 1
#
Iterator: 2

Iterator: 3

Iterator: 4

Iterator: 5

Iterator: 6

Iterator: 7

PRINT inputString
-----
0
Segmentation fault: 11

我的问题:

1. 为什么 argv 包含 3 个以上的元素(M、# 和\0)。它打印出 8 个元素(打印语句迭代 0 - 7)，这些元素在 W、#、\0 之后是垃圾。它不应该只打印 3 个元素(M、# 和\0)吗？为什么会这样？我该如何解决？
2. 为什么当我将 newString(char 类型)设置为 inputString(char* 类型)时，通过执行 inputString = newString，inputString 在 print 语句中迭代 8 次，在打印 M、# 和\0 之后打印空白。
3. 为什么第三条语句会出现seg fault？

Answer 1

sizeof() 不返回空终止字符数组字符串的长度。相反，您需要类似 strlen() 的东西。