我对Python编码非常熟悉，但现在我必须在C语言中进行字符串分析。
我的意见：
input =“command1 args1 args2 arg3; command2 args1 args2 args3; cmd3 arg1 arg2 arg3”
我的Python解决方案：
input = "command1 args1 args2 arg3;command2 args1 args2 args3;command3 arg1 arg2 arg3"
compl = input.split(";")
tmplist =[]
tmpdict = {}
for line in compl:
spl = line.split()
tmplist.append(spl)
for l in tmplist:
first, rest = l[0], l[1:]
tmpdict[first] = ' '.join(rest)
print tmpdict
#The Output:
#{'command1': 'args1 args2 arg3', 'command2': 'args1 args2 args3', 'cmd3': 'arg1 arg2 arg3'}
预期输出：使用命令作为键进行Dict，并将args作为值
中的字符串连接 到目前为止我的C解决方案：
我想将命令和args保存在这样的结构中：
struct cmdr{
char* command;
char* args[19];
};
我创建一个struct char *数组来保存由“;”分隔的cmd + args：
struct ari {char * value [200];};
功能：
struct ari inputParser(char* string){
char delimiter[] = ";";
char *ptrsemi;
int i = 0;
struct ari sepcmds;
ptrsemi = strtok(string, delimiter);
while(ptrsemi != NULL) {
sepcmds.value[i] = ptrsemi;
ptrsemi = strtok(NULL, delimiter);
i++;
}
return sepcmds;
首先我添加了一个帮助结构：
struct arraycmd {
struct cmdr lol[10];
};
struct arraycmd parseargs (struct ari z){
struct arraycmd result;
char * pch;
int i;
int j = 0;
for (i=0; i < 200;i++){
j = 0;
if (z.value[i] == NULL){
break;
}
pch = strtok(z.value[i]," ");
while(pch != NULL) {
if (j == 0){
result.lol[i].command = pch;
pch = strtok(NULL, " ");
j++;
} else {
result.lol[i].args[j]= pch;
pch = strtok(NULL, " ");
j++;
}
}
pch = strtok(NULL, " ");
}
return result;
我的输出功能如下：
void output(struct arraycmd b){
int i;
int j;
for(i=0; i<200;i++){
if (b.lol[i].command != NULL){
printf("Command %d: %s",i,b.lol[i].command);
}
for (j = 0; j < 200;j++){
if (b.lol[i].args[j] != NULL){
printf(" Arg %d = %s",j,b.lol[i].args[j]);
}
}
printf(" \n");
}
}
但它只生成垃圾（与我的python解决方案中相同的输入）：
（command1 args1 args2 arg3; command2 args1 args2 args3; command3 arg1 arg2 arg3）
命令0：command1 Arg 0 = command2 Arg 1 = args1 Arg 2 = args2 Arg 3 = arg3 Arg 19 = command2 Arg 21 = args1 Arg 22 = args2 Arg 23 = args3 Arg 39 = command3 Arg 41 = arg1 Arg 42 = arg2 Arg 43 = arg3 分段错误
所以我希望有人可以帮我解决这个问题。
答案 0 :(得分：1)
在python中直接获得C逻辑可能更容易。这更接近于C，您可以尝试将其音译为C.您可以使用strncpy来提取字符串并将其复制到您的结构中。
str = "command1 args1 args2 arg3;command2 args1 args2 args3;command3 arg1 arg2 arg3\000"
start = 0
state = 'in_command'
structs = []
command = ''
args = []
for i in xrange(len(str)):
ch = str[i]
if ch == ' ' or ch == ';' or ch == '\0':
if state == 'in_command':
command = str[start:i]
elif state == 'in_args':
arg = str[start:i]
args.append(arg)
state = 'in_args'
start = i + 1
if ch == ';' or ch == '\0':
state = 'in_command'
structs.append((command, args))
command = ''
args = []
for s in structs:
print s
答案 1 :(得分：1)
您的问题是您依赖结构中的指针初始化为NULL。
它们只是随机值，因此是SEGV。
当结构只有10个命令和19个参数时，您还打印200个命令和200个参数。
答案 2 :(得分：1)
检查此解决方案。用valgrind测试没有泄漏。 但是我在内部实现打印.u可以自己实现查看自由函数。另外你可以改进分割函数来实现更好的解析。
#include <stdio.h>
#include <stdlib.h>
typedef struct arr {
char** words;
int count;
} uarr;
#define null 0
typedef struct cmdr {
char* command;
char** argv;
int argc;
} cmd;
typedef struct list {
cmd* listcmd;
int count;
} cmdlist;
uarr splitter(char* str, char delim);
cmdlist* getcommandstruct(char* string);
void freecmdlist(cmdlist* cmdl);
int main(int argc, char** argv) {
char input[] = "command1 arg1 arg2 arg3 arg4;command2 arg1 arg2 ;command3 arg1 arg2 arg3;command4 arg1 arg2 arg3";
cmdlist* cmdl = getcommandstruct((char*) input);
//it will free . also i added print logic inside free u can seperate
freecmdlist(cmdl);
free(cmdl);
return (EXIT_SUCCESS);
}
/**
* THIS FUNCTION U CAN USE FOR GETTING STRUCT
* @param string
* @return
*/
cmdlist* getcommandstruct(char* string) {
cmdlist* cmds = null;
cmd* listcmd = null;
uarr resultx = splitter(string, ';');
//lets allocate
if (resultx.count > 0) {
listcmd = (cmd*) malloc(sizeof (cmd) * resultx.count);
memset(listcmd, 0, sizeof (cmd) * resultx.count);
int i = 0;
for (i = 0; i < resultx.count; i++) {
if (resultx.words[i] != null) {
printf("%s\n", resultx.words[i]);
char* def = resultx.words[i];
uarr defres = splitter(def, ' ');
listcmd[i].argc = defres.count - 1;
listcmd[i].command = defres.words[0];
if (defres.count > 1) {
listcmd[i].argv = (char**) malloc(sizeof (char*) *(defres.count - 1));
int j = 0;
for (; j < defres.count - 1; j++) {
listcmd[i].argv[j] = defres.words[j + 1];
}
}
free(defres.words);
free(def);
}
}
cmds = (cmdlist*) malloc(sizeof (cmdlist));
cmds->count = resultx.count;
cmds->listcmd = listcmd;
}
free(resultx.words);
return cmds;
}
uarr splitter(char* str, char delim) {
char* holder = str;
uarr result = {null, 0};
int count = 0;
while (1) {
if (*holder == delim) {
count++;
}
if (*holder == '\0') {
count++;
break;
};
holder++;
}
if (count > 0) {
char** arr = (char**) malloc(sizeof (char*) *count);
result.words = arr;
result.count = count;
//real split
holder = str;
char* begin = holder;
int index = 0;
while (index < count) {
if (*holder == delim || *holder == '\0') {
int size = holder + 1 - begin;
if (size > 1) {
char* dest = (char*) malloc(size);
memcpy(dest, begin, size);
dest[size - 1] = '\0';
arr[index] = dest;
} else {
arr[index] = null;
}
index++;
begin = holder + 1;
}
holder++;
}
}
return result;
}
void freecmdlist(cmdlist* cmdl) {
if (cmdl != null) {
int i = 0;
for (; i < cmdl->count; i++) {
cmd def = cmdl->listcmd[i];
char* defcommand = def.command;
char** defargv = def.argv;
if (defcommand != null)printf("command=%s\n", defcommand);
free(defcommand);
int j = 0;
for (; j < def.argc; j++) {
char* defa = defargv[j];
if (defa != null)printf("arg[%i] = %s\n", j, defa);
free(defa);
}
free(defargv);
}
free(cmdl->listcmd);
}
}