C代码中使用演示参数的问题确实处理了argv中的字符串

Question

我有一个C代码显示一个参数的小写字母：从命令行使用超过6个参数调用时，此代码可以正常运行但是，如果不带参数调用，演示部分将无法正常工作，并且程序将卡住：

下面是代码文件：

谁能帮我，谢谢！

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

int showLowerCase(char *argv[]){
            char* aStr = argv[6];
            //To lowercase: the demo got stuck probably here
            for (int i = 0; aStr[i]; i++) {
                aStr[i] = tolower(aStr[i]);
            }
            printf("a lower case for 6th input string: %s\n",aStr);
}

int main(int argc, char *argv[])
{
        if(argc < 7){
            // when no srguments given, try a demo:
            // change argc to 7
            argc = 7;
            // make a example of argv2 with 7 strings
            char *argv2[7];
            argv2[0] = "killWindowsVersatile.exe";
            argv2[1] = "key";
            argv2[2] = "ci";
            argv2[3] = "once";
            argv2[4] = "2";
            argv2[5] = "1000";
            argv2[6] = "SuperCol"; //this argument will be shown as lowercase
            showLowerCase(argv2);
            return 0;
        }
        showLowerCase(argv);
        return 0;
}

似乎argv可以在现场修改为小写，而我构建的argv2无法修改。

Answer 1

由strcpy解决：

int showLowerCase(char *argv[]){
            printf("aStr = argv[6]\n");
            char* aStr = argv[6];
            printf("aStr = argv[6]...done");
            //make hard copy
            char newStr[strlen(aStr)+1];
            strcpy(newStr,aStr);

            //To lowercase
            for (int i = 0; newStr[i]; i++) {
                printf("i=0\n");
                printf("%c\n",newStr[i]);
                newStr[i] = tolower(aStr[i]);
            }
            printf("a lower case for 6th input string: %s\n",newStr);
}