问题描述
关于将2D数组转换为1D数组有很多问题，但我正尝试相反的尝试。我正在尝试将一个字符串划分为恒定长度的子字符串，并将其放置在2D数组中。此2D矩阵的每一行都应包含初始字符串的子字符串，并且，如果要连续读取并连接每一行，则应复制初始字符串。
There are a lot of questions about converting a 2D array into a 1D array, but I am attempting just the opposite. I'm trying to partition a string into substrings of constant length and house them in a 2D array. Each row of this 2D matrix should contain a substring of the initial string, and, if each row were to be read in succession and concatenated, the initial string should be reproduced.
我几乎可以使用它了，但是由于某种原因，我丢失了初始字符串（bin）的第一个子字符串（分区[0]-长度8 * blockSize）：
I nearly have it working, but for some reason I am losing the first substring (partitions[0] -- length 8*blockSize) of the initial string (bin):
int main (void){
char* bin = "00011101010000100001111101001101000010110000111100000010000111110100111100010011010011100011110000011010";
int blockSize = 2; // block size in bytes
int numBlocks = strlen(bin)/(8*blockSize); // number of block to analyze
char** partitions = (char**)malloc((numBlocks+1)*sizeof(char)); // break text into block
for(int i = 0; i<numBlocks;++i){
partitions[i] = (char*)malloc((8*blockSize+1)*sizeof(char));
memcpy(partitions[i],&bin[8*i*blockSize],8*blockSize);
partitions[i][8*blockSize] = '\0';
printf("Printing partitions[%d]: %s\n", i, partitions[i]);
}
for(int j=0; j<numBlocks;++j)
printf("Printing partitions[%d]: %s\n", j,partitions[j]);
return 0;
}
输出如下：
Printing partitions[0]: 0001110101000010
Printing partitions[1]: 0001111101001101
Printing partitions[2]: 0000101100001111
Printing partitions[3]: 0000001000011111
Printing partitions[4]: 0100111100010011
Printing partitions[5]: 0100111000111100
Printing partitions[0]: Hj
Printing partitions[1]: 0001111101001101
Printing partitions[2]: 0000101100001111
Printing partitions[3]: 0000001000011111
Printing partitions[4]: 0100111100010011
Printing partitions[5]: 0100111000111100
在第一个for循环中成功构建分区。读取后进行构造后，partitions [0]处的字符串包含垃圾值。谁能提供一些见识？
The construction of partitions in the first for loop is successful. After construction at read out, the string at partitions[0] contains garbage values. Can anyone offer some insight?
推荐答案
他们的代码存在一些问题。
Their are a few problems with your code.
-
您分配的
**分区错误。
而不是：
char** partitions = (char**)malloc((numBlocks+1)*sizeof(char)); /* dont need +1, as numblocks is enough space. */
您需要为 char * 指针，而不是 char 个字符。
You need to allocate space for char* pointers, not char characters.
相反，它需要是：
char** partitions = malloc((numBlocks+1)*sizeof(char*));
也请阅读，因为在C语言中不需要。
Also read Why not to cast result of malloc(), as it is not needed in C.
malloc（），因为不成功时它可能返回 NULL 。
malloc() needs to be checked everytime, as it can return NULL when unsuccessful.
以下代码显示了此内容：
Here is some code which shows this:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define BLOCKSIZE 2
#define BLOCK_MULTIPLIER 8
int main(void) {
const char *bin = "00011101010000100001111101001101000010110000111100000010000111110100111100010011010011100011110000011010";
const size_t blocksize = BLOCKSIZE;
const size_t multiplier = BLOCK_MULTIPLIER;
const size_t numblocks = strlen(bin)/(multiplier * blocksize);
const size_t numbytes = multiplier * blocksize;
char **partitions = malloc(numblocks * sizeof(*partitions));
if (partitions == NULL) {
printf("Cannot allocate %zu spaces\n", numblocks);
exit(EXIT_FAILURE);
}
for (size_t i = 0; i < numblocks; i++) {
partitions[i] = malloc(numbytes+1);
if (partitions[i] == NULL) {
printf("Cannot allocate %zu bytes for pointer\n", numbytes+1);
exit(EXIT_FAILURE);
}
memcpy(partitions[i], &bin[numbytes * i], numbytes);
partitions[i][numbytes] = '\0';
printf("Printing partitions[%zu]: %s\n", i, partitions[i]);
}
printf("\n");
for(size_t j = 0; j < numblocks; j++) {
printf("Printing partitions[%zu]: %s\n", j,partitions[j]);
free(partitions[j]);
partitions[j] = NULL;
}
free(partitions);
partitions = NULL;
return 0;
}
输出非垃圾值：
Printing partitions[0]: 0001110101000010
Printing partitions[1]: 0001111101001101
Printing partitions[2]: 0000101100001111
Printing partitions[3]: 0000001000011111
Printing partitions[4]: 0100111100010011
Printing partitions[5]: 0100111000111100
Printing partitions[0]: 0001110101000010
Printing partitions[1]: 0001111101001101
Printing partitions[2]: 0000101100001111
Printing partitions[3]: 0000001000011111
Printing partitions[4]: 0100111100010011
Printing partitions[5]: 0100111000111100
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