指针失去其价值+ execv编译警告

问题描述

我希望我没有错过过类似的问题。

I hope I haven't missed a similar question.

我想code我自己的一个小型外壳，采用原始的C函数。

I'm trying to code a mini-shell of my own, using primitive C functions.

我得到的东西应该工作，但我有一个指针，使一切错误。

I got something that should work, but I have a pointer that makes everything bug.

我的 adrCmd 指针应该从 searchCmd（）函数命令路径字符串，并保持相同的值在主功能。

My adrCmd pointer should get the command-path string from the searchCmd() function and keep the same value in the main function.

在的事实：它指向右边值 searchCmd（），但不是在的main（）

In fact: it points to the right value on searchCmd(), but not in the main().

这里的code：

    int searchCmd(char* cmd, char* adrCmd){
    char* path = getenv("PATH");
    if(debug)printf("PATH : %s\n", path);
    int nbPath = (compteLettre(path, ':')+1);
    char** pathTab = malloc(nbPath*sizeof(char*));
    decompose(path, pathTab, 2048, ':');

    int i;
    char* adr = malloc(sizeof(char*));
    for(i=0; i<nbPath; i++){
        sprintf(adr, "%s/%s", pathTab[i], cmd);
        if(debug)printf("   source : %s \n", adr);


        int fs = open(adr, O_RDONLY); // Si on peut ouvrir le fichier, c'est qu'il existe !
        if(fs != -1){ // si le fichier existe, on le renvoie;
            if(debug){
                printf("Commande trouvée dans path ! \n");
                printf("%s \n", adr);
            }
            adrCmd = adr;
            printf("%s ?= %s \n",adrCmd, adr );// oui
            return 1;
        }
    }
    return 0;
}

/**********************\
        Main
\**********************/

int main(int argc, char** argv){
    printf("Mini-shell : OK \n");

    char cmd[CMDSIZE];
    char** splited = malloc(CMDSIZE*sizeof(char*));
    char* adrCmd = malloc(sizeof(char*));
    char* params;
    while(printf("$ : ") && gets(cmd) && (strcmp(cmd, "exit")!=0 && strcmp(cmd, "quit")!=0)){ // On boucle tant que la commande != "exit" ou "quit"

        printf("Votre commande : %s \n", cmd);
        decompose(cmd, splited, CMDSIZE, ' ');
        if(debug)afficheCmd(splited, CMDSIZE);

        if(!searchCmd(splited[0], adrCmd)){
            printf("Commande n'existe pas, essayez apt-get install %s\n", splited[0]);
        }else{
            if(debug)printf("Execution de la commande '%s' : \n", adrCmd);
            params = splited[1]; // params = array(splited[1], splited[2], ...... )
            if(execv(adrCmd, params) == -1){
                printf("Erreur d'exection de la commande\n");
            }
        }

    }
    printf("Fin du programme %s \n", argv[0]);
    return 0;
}

下面是执行返回的内容：

Here's what the execution returns:

$ ./a.out
Mini-shell : OK
$ : ls /var
Votre commande : ls /var
CMD[0] = ls
CMD[1] = /var
PATH : /usr/lib/lightdm/lightdm:/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin:/usr/games
    source : /usr/lib/lightdm/lightdm/ls
    source : /usr/local/sbin/ls
    source : /usr/local/bin/ls
    source : /usr/sbin/ls
    source : /usr/bin/ls
    source : /sbin/ls
    source : /bin/ls
Commande trouvée dans path !
/bin/ls
/bin/ls ?= /bin/ls
Execution de la commande '' :
Erreur d'exection de la commande

虽然我在这里，当我编译， execv 返回警告：

$ gcc shell.c
shell.c: In function ‘main’:
shell.c:113:4: attention : passing argument 2 of ‘execv’ from incompatible pointer type [enabled by default]
/usr/include/unistd.h:564:12: note: expected ‘char * const*’ but argument is of type ‘char *’

我应该怎么做才能避免这种情况？

What should I do to avoid this?

推荐答案

C是传值。因此，执行此操作时

C is pass by value. So when doing this

 int searchCmd(char * cmd, char * adrCmd){

adrCmd 是怎样在了已经通过的副本。覆盖副本不会改变什么已经从来电者复制。

adrCmd is a copy of what had been passed in. Overwriting the copy won't change what it had been copied from in the caller.

要解决这个向下传递的地址 adrCmd ：

To fix this pass down the address of adrCmd:

 int searchCmd(char * cmd, char ** padrCmd){

和使用这样的：

    *padrCmd = adr;

呼叫 searchCmd（）是这样的：

  if(!searchCmd(splited[0], &adrCmd)){

和定义和初始化 adrCmd 这样的;

and define and initialise adrCmd like this;

  char * adrCmd = NULL;

这篇关于指针失去其价值+ execv编译警告的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！