将标记从 1D char 数组存储到 char** 数组

Question

我正在尝试编写一个程序，该程序将动态分配足够的 space 以将所有单词存储在由 space 分隔的一维字符数组中。例如：

char *literal = "The quick brown fox";
char **words = { "The", "quick", "brown", "fox" };

我写的程序在尝试 strncpy(str[buff_ptr],tok,strlen(tok));

时一直出现段错误

我将 post 我的代码如下：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char *mutableString(char *lit) {
    int size = strlen(lit);
    char *str = (char *)malloc(sizeof(char) * size);
    strncpy(str, lit, size + 1);
    return str;
}

int numTokens(char *str, const char *DELIM) {
    char* clone = (char*)malloc(sizeof(char*));
    strncpy(clone, str, strlen(str) + 1);
    int count = 0;
    for (char *tok = strtok(clone, " "); tok != NULL; tok = strtok(NULL, " "))
        count++;
    free(clone);
    return count;
}

char **tokenize(char *str, const char *DELIM) {
    printf("tokenize-------------------------\n");
    int size = numTokens(str, DELIM);
    //allocate space on heap for buffer
    char **buff = (char **)malloc(size * sizeof(char *));
    //get first word
    char *tok = strtok(str, DELIM);
    int buff_ptr = 0;
    while (tok != NULL) {
        strncpy(buff[buff_ptr], tok, strlen(tok) + 1);
        printf("buff[%d]%s\n", buff_ptr, buff[buff_ptr]);
        //increment to next word for storage
        buff_ptr++;
        //find next word in string
        tok = strtok(NULL, DELIM);
    }
    for (int i = 0; i < size; i++) {
        printf("%s\n", buff[i]);
    }
    //return 2D pointer
    return buff;
}

int main() {
    char *literal = "some literal string.";
    //convert string to mutable string for strtok
    char *str = mutableString(literal);
    //set 2D pointer equal to the pointer address returned
    char **no_spaces_str = tokenize(str, " ");
    printf("%s\n", str);
    for (int i = 0; i < numTokens(str, " "); i++) {
        printf("%s\n", no_spaces_str[i]);
    }
    //free heap allocated memory
    free(str);
    free(no_spaces_str);
    return 0;
}

请参阅lldb堆栈变量附件：

Answer 1

在函数 mutableString 中动态分配了不包含字符串的字符数组 str

char* mutableString(char* lit){
  int size = strlen(lit);
  char* str = (char*)malloc(sizeof(char)*size);
  strncpy(str,lit,size);
  return str;
}

所以其他函数调用未定义的行为，例如在这个 for 循环中

int numTokens(char* str, const char* DELIM){
  int count = 0;
  for(; *str != '[=11=]'; str++)
  //...

此外，如果数组包含一个字符串，那么函数 numTokens 是不正确的，因为例如它 returns 0 当传递的字符串仅包含一个单词时。

也在函数tokenize

strncpy(buff[buff_ptr],tok,strlen(tok));

使用了未初始化的指针 buff[buff_ptr] 分配类似。

char **buff = (char**)malloc(size*sizeof(char*));

您再次尝试复制字符串而不包括终止零字符 '\0;使用 eth 函数 strncpy.

所以这个调用在 main

printf("%s\n",no_spaces_str[i]);

也将调用未定义的行为。

Answer 2

除了提到的@Vlad from Moscow点之外，

malloc return 值不能是 type-casted Do I cast the result of malloc?

我试图清理代码找到下面的片段，DEMO

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct{
    char** buff;
    int    size;
}Array_2d;

char* mutableString(const char* lit){
  int size = strlen(lit);
  char* str = malloc(size);
  strncpy(str,lit,size+1);
  return str;
}

int getNextWordLength(const char* str){
  int index = 0;
  while(*str && (*str != ' ')){
      //printf("%c",*str);
      ++index;
      ++str;
  }
  return index;
}

int numTokens(const char* str){
  int count = 0;
  for(; *str != '[=10=]'; str++)
  {
    if(*str == ' ')
      count++;
  }
  return count;
}

void tokenize(const char* str, const char *DELIM, Array_2d *array){

  int len = strlen(str)+1;

    if(!str && !len){
        array->buff = 0;
        array->size = 0;
    }

    int number_of_words = numTokens(str)+1;
    //allocate space on heap for buffer
    char **buff = (char**)malloc(number_of_words*sizeof(char*));

    int index = 0; 

    do{
        //get first word
        
        int word_length = getNextWordLength(str);
        
        //To compensate null terminal
        buff[index] = malloc(word_length+1);

        strncpy(buff[index], str,word_length);

        buff[index][word_length+1] = '[=10=]';

        str += word_length+1;
        ++index;

    }while(index < number_of_words);

    //update return value
    array->buff = buff;
    array->size = number_of_words;

}

int main(){
    char* literal = "hello world this is test";
    //convert string to mutatable string for strtok
    char* str = mutableString(literal);
    printf("Complete String is : %s\n",str);
    

    Array_2d array;
    // set 2D pointer equal to the pointer addres returned
     tokenize(str, " ",&array);

    printf("Tokenized String\n");
    for(int i=0;i<array.size;i++){
        printf("%s\n",array.buff[i]);
    }

    free(str);

    for(int i =0;i< array.size; ++i)
    free(array.buff[i]);

    free(array.buff);

  return 0;
}

Answer 3

这是上面代码的更正版本

当你复制字符串时你应该为 '\0' 添加 1 个字符

int size = strlen(lit)+1;
令牌缓冲区大小应为大小+1

int size = numTokens(str, DELIM)+1;
不需要 Strncpy strncpy(buff[buff_ptr], tok, strlen(tok) + 1); 你已经复制了字符串 char* str = mutableString(literal); 只需指向每个下一个标记的第 n 个缓冲区 buff[buff_ptr]=tok;
```
for (int i = 0; i<numTokens(str, " "); i++){
    printf("%s\n", no_spaces_str[i]);
}
```
此代码无法正常工作。 strtok 操作您传入的字符串和 returns 指向它的指针，因此不会分配内存。所以所有空格都将替换为 '\0'

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#pragma warning(push)
#pragma warning(disable : 4996)

char* mutableString(char* lit){
    int size = strlen(lit)+1;
    char* str = (char*)malloc(sizeof(char)*size);
    strncpy(str, lit, size);
    return str;
}
int numTokens(char* str, const char* DELIM){
    int count = 0;
    for (; *str != '[=11=]'; str++)
    {
        if (*str == ' ')
            count++;
    }
    return count;
}
char** tokenize(char* str, const char* DELIM){
    printf("tokenize-------------------------\n");
    int size = numTokens(str, DELIM)+1;
    //allocate space on heap for buffer
    char **buff = (char**)malloc((size)*sizeof(char*));
    //get first word
    char* tok = strtok(str, DELIM);
    int buff_ptr = 0;
    while (tok != NULL){

        buff[buff_ptr]=tok;
        printf("buff[%d]%s\n", buff_ptr, buff[buff_ptr]);
        //increment to next word for storage
        buff_ptr++;
        //find next word in string
        tok = strtok(NULL, DELIM);
    }

    for (int i = 0; i<size; i++){
        printf("%s\n", buff[i]);
    }
    //return 2D pointer
    return buff;
}
int main(){
    char* literal = "some literal string.";
    //convert string to mutatable string for strtok
    char* str = mutableString(literal);
    //set 2D pointer equal to the pointer addres returned
    char** no_spaces_str = tokenize(str, " ");
    printf("%s\n", str);

    for (int i = 0; i<numTokens(str, " "); i++){
        printf("%s\n", no_spaces_str[i]);
    }
    //free heap allocated memory
    free(str);

    free(no_spaces_str);
    return 0;
}

结果

tokenize-------------------------
buff[0]some
buff[1]literal
buff[2]string.
some
literal
string.
some

Answer 4

char* mutableString(char* lit){
  int size = strlen(lit)+1;
  char* str = (char*)malloc(sizeof(char)*size);
  strncpy(str,lit,size);
  return str;
}
int numTokens(char* str, const char* DELIM){
  int size = strlen(str)+1;
  char* clone = (char*)malloc(sizeof(char)*size);
  strncpy(clone,str,size);
  int count = 0;
  for(char* tok = strtok(clone," "); tok != NULL; tok=strtok(NULL, " "))
      count++;
  free(clone);
  return count;
}
char** tokenize(char* str, const char* DELIM){
  int size = strlen(str)+1;
  char* clone = (char*)malloc(sizeof(char)*size);
  strncpy(clone,str,size);
  // printf("tokenize-------------------------\n");
  int size = numTokens(str, DELIM);
  //allocate space on heap for buffer
  char **buff = (char**)calloc(size,sizeof(char*));
  //get first word
  char* tok = strtok(clone,DELIM);
  int buff_ptr = 0;
  while(tok != NULL){
    // printf("token%d:%s\n",buff_ptr,tok);
    buff[buff_ptr] = (char*)malloc(sizeof(char)*strlen(tok)+1);
    strncpy(buff[buff_ptr],tok,strlen(tok)+1);
    //increment to next word for storage
    buff_ptr++;
    //find next word in string
    tok = strtok(NULL, DELIM);
  }
  //return 2D pointer
  free(clone);
  return buff;
}
int main(){
  char* literal = "some literal string.";
  //convert string to mutatable string for strtok
  char* str = mutableString(literal);
  //set 2D pointer equal to the pointer addres returned
  char** no_spaces_str = tokenize(str, " ");
  int num_words = numTokens(str," ");
  char* oneD = (char*)calloc(strlen(str)+1,sizeof(char));
  for(int i = 0;i<num_words;i++){
    strncat(oneD,no_spaces_str[i],strlen(no_spaces_str[i])+1);
    printf("%s\n",oneD);
  }
  
  //free heap allocated memory
  free(str);
  free(no_spaces_str);
  free(oneD);
  return 0;
}

可以解决我的问题。感谢所有评论并帮助我更好地理解动态内存的人。

将标记从 1D char 数组存储到 char** 数组

Storing tokens from 1D char array to char** array

c

split

c-strings

strncpy

dynamic-memory-allocation