如何遍历char **变量

Question

我正在运行一个小型rpc程序/使用一个rpc框架，该框架从客户端文件中获取一个char[]，并将其发送到枚举字符串中整数的服务器。

我有一个文件client.c，它接受用户输入并将其传递给头文件中的外部函数。

#include <rpc/rpc.h>
#include "getNumInt.h"

int main(int argc, char **argv){
    CLIENT *cli;
    char *server;
    server = argv[1];

    cli = clnt_create(server, GETNUMINT, GNUMINT, "tcp"); //creates a client handle
    /*does some check for whether the client connected*/
    char command[256];
    int *numInt;
    fgets(command, 256, stdin); 
    numInt = enumints_1(&command, cli); //segfaults here according to backtrace
    return(0);
}

函数enumints_1在我的服务器存根server.c中定义为：

int *enumints_1(msg, req)
    char **msg; struct svc_req *req;{
    printf(*msg);
    static int numDigits = 0;
    char msgcopy[256];
    strcpy(msgcopy, *msg);
    int i = 0;
    for(i; i<strlen(msgcopy); i++){
        if(msgcopy[i] >= '0' && msgcopy[i] <='9'){
            numDigits++;
        }
    }
    return(&numDigits);
}

我的主要问题是如何迭代char **msg，因为这很可能是我的程序出现分段错误的原因。command只是从用户输入中取出的字符串，然后通过引用传递给enumints_1函数。因为它是指向某个点的指针，所以我假设我可以只使用strcpy或memcpy将字符串复制到char数组，但这是行不通的。

也是我的.x文件：

struct intStringPair{
    int numInts;
    char msg[256];
};

program GETNUMINT{
    version GNUMINT{
        int ENUMINTS(string) = 1; //string considered char * the rpc generated file makes it so enumints_1 then has to take char **
        int WRITEMESSAGE(intStringPair) = 2;
    } = 1;
}= 0x20000001;

Answer 1

类似@said 3386109这样说：

命令不是指针。所以&命令不是指针到指针。

因此，将&命令分配给msg是无效的(而且我的编译器甚至都不编译它)

当我编译这段代码时：

// my set up code

#include <stdio.h>
#include <string.h>
struct svc_req {};
typedef struct svc_req CLIENT;
struct svc_req *clnt_create(const char *, int, int, const char*) {return 0;}
int GETNUMINT=0, GNUMINT=0;

// your code verbatim

int *enumints_1(char **msg, struct svc_req *req){
    printf(*msg);
    static int numDigits = 0;
    char msgcopy[256];
    strcpy(msgcopy, *msg);
    int i = 0;
    for(i; i<strlen(msgcopy); i++){
        if(msgcopy[i] >= '0' && msgcopy[i] <='9'){
            numDigits++;
        }
    }
    return(&numDigits);
}

int main(int argc, char **argv){
    CLIENT *cli;
    char *server;
    server = argv[1];

    cli = clnt_create(server, GETNUMINT, GNUMINT, "tcp"); //creates a client handle
    /*does some check for whether the client connected*/
    char command[256];
    int *numInt;
    fgets(command, 256, stdin); 
    numInt = enumints_1(&command, cli); //segfaults here according to backtrace
    return(0);
}

编译器说：

<source>: In function 'int main(int, char**)':
<source>:34:25: error: cannot convert 'char (*)[256]' to 'char**'
     numInt = enumints_1(&command, cli); //segfaults here according to backtrace
                         ^~~~~~~~
<source>:10:24: note:   initializing argument 1 of 'int* enumints_1(char**, svc_req*)'
 int *enumints_1(char **msg, struct svc_req *req){
                 ~~~~~~~^~~
Compiler returned: 1

相反，您可以做的是创建一个指向数组的指针，然后传递该数组的地址：

// my set up code
#include <stdio.h>
#include <string.h>
struct svc_req {};
typedef struct svc_req CLIENT;
struct svc_req *clnt_create(const char *, int, int, const char*) {return 0;}
int GETNUMINT=0, GNUMINT=0;

// your code verbatim
int *enumints_1(char **msg, struct svc_req *req){
    printf(*msg);
    static int numDigits = 0;
    char msgcopy[256];
    strcpy(msgcopy, *msg);
    int i = 0;
    for(i; i<strlen(msgcopy); i++){
        if(msgcopy[i] >= '0' && msgcopy[i] <='9'){
            numDigits++;
        }
    }
    return(&numDigits);
}

int main(int argc, char **argv){
    CLIENT *cli;
    char *server;
    server = argv[1];

    cli = clnt_create(server, GETNUMINT, GNUMINT, "tcp"); //creates a client handle
    /*does some check for whether the client connected*/
    char command[256], *command_pointer=command;
    int *numInt;
    fgets(command, 256, stdin); 
    numInt = enumints_1(&command_pointer, cli); //segfaults here according to backtrace
    return(0);
}