在C++中操作动态分配的二维char数组
在C++中操作动态分配的二维char数组
提问于 2019-02-18 23:51:39
我在C中尝试操作2d动态数组时遇到了麻烦，我想做的是在2d数组的每一行中存储一个char字符串，然后执行一个检查，看看字符串是否包含某个特定的字符，如果包含，则删除所有出现的字符，然后在空位上移动。实际发生的情况是我得到了一个exit status 1。
更多关于问题的信息，例如，如果我有
Enter string 1: testing
Enter string 2: apple
Enter string 3: banana我希望输出成为
What letter? a // ask what character to search for and remove all occurences
testing
pple
bnn下面是我的完整代码：
#include <stdio.h>
#include <stdlib.h>
void removeOccurences2(char** letters, int strs, int size, char letter){
// Get size of array
// Shift amount says how many of the letter that we have removed so far.
int shiftAmt = 0;
// Shift array says how much we should shift each element at the end
int shiftArray[strs][size];
// The first loop to remove letters and put things the shift amount in the array
int i,j;
for(i=0;i < strs; i++){
for(j = 0; j < size - 1; j++) {
if (letters[i][j] == '\0'){
break;
}
else {
// If the letter matches
if(letter == letters[i][j]){
// Set to null terminator
letters[i][j] = '\0';
// Increase Shift amount
shiftAmt++;
// Set shift amount for this position to be 0
shiftArray[i][j] = 0;
}else{
// Set the shift amount for this letter to be equal to the current shift amount
shiftArray[i][j] = shiftAmt;
}
}
}
}
// Loop back through and shift each index the required amount
for(i = 0; i < strs; i++){
for(j = 0; j < size - 1; j++) {
// If the shift amount for this index is 0 don't do anything
if(shiftArray[i][j] == 0) continue;
// Otherwise swap
letters[i][j - shiftArray[i][j]] = letters[i][j];
letters[i][j] = '\0';
}
//now print the new string
printf("%s", letters[i]);
}
return;
}
int main() {
int strs;
char** array2;
int size;
int cnt;
int c;
char letter;
printf("How many strings do you want to enter?\n");
scanf("%d", &strs);
printf("What is the max size of the strings?\n");
scanf("%d", &size);
array2 = malloc(sizeof(char*)*strs);
cnt = 0;
while (cnt < strs) {
c = 0;
printf("Enter string %d:\n", cnt + 1);
array2[cnt] = malloc(sizeof(char)*size);
scanf("%s", array2[cnt]);
cnt += 1;
}
printf("What letter?\n");
scanf(" %c", &letter);
removeOccurences2(array2,strs,size,letter);
}提前感谢！
回答 1
Stack Overflow用户
回答已采纳
发布于 2019-02-19 01:26:16
您可以就地删除字符串中的字母，因为您只能缩短字符串。
代码可以简单地为：
void removeOccurences2(char** letters, int strs, int size, char letter){
int i,j,k;
// loop over the array of strings
for(i=0;i < strs; i++){
// loop per string
for(j = 0, k=0; j < size; j++) {
// stop on the first null character
if (letters[i][j] == '\0'){
letters[i][k] = 0;
break;
}
// If the letter does not match, keep the letter
if(letter != letters[i][j]){
letters[i][k++] = letters[i][j];
}
}
//now print the new string
printf("%s\n", letters[i]);
}
return;
}但是您应该在返回环境之前释放所有已分配的数组，并在main的末尾显式返回0。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/54750958
复制相关文章
相似问题