我正在尝试编写一个程序，如果以特定格式输入，则将单词分解成段/音节。我想以格式输入一个单词：
[d-i-s][c-o-#][v-er-#]
并将其存储为以下格式：
syllable[0]=dis
syllable[1]=co
syllable[2]=ver
到目前为止，我已设法使用分隔符']'将其分解为音节，并将其保存为以下格式的char **：
syllable[0]=[d-i-s
syllable[1]=[c-o-#
syllable[2]=[v-er-#
所以现在我只想清理它并删除不必要的角色！我以为我会创建一个新数组并复制旧数组中的字母，只要它们不是[ - ＃。但是对于我的生活，我无法弄清楚如何将正确的字母复制到另一个阵列中!!
我知道我不能这样做：
cleanArray[i][j] = dirtyArray[i][k]
因为cleanArray [i]是char *而我无法编辑它？但可以我做什么？
我已经阅读了很多类似的问题，这些问题建议使用strncpy和snprintf（how to copy char array to another char array in C?，strcpy and printf a multidimensional char array C），但我已尝试过这些问题但我无法让它们发挥作用。我甚至尝试将cleanArray放入3个维度，希望我能够将单个字母保存为cleanArray [i] [j]作为char * s，这可能是完全错误的。
正确的方法是什么？对不起，如果它很明显，但我花了几个小时，现在非常，非常，困惑......我真的很感激你能给出的任何建议！
这是我的代码：
char** cleanStrings (char**dirtyList, int arrayLength)
{
int i, j, k;
char** cleanList = (char**)calloc(arrayLength, CHARLEN);
for (i=0; i<arrayLength; i++)
{
k= 0;
cleanList[i] = (char*)calloc(10,CHARLEN);
for (j=0; j<strlen(dirtyList[i]+1);j++)
{
if (dirtyList[i][j] == '[') continue;
else if (dirtyList[i][j] == '#') continue;
else if (dirtyList[i][j] == '-') continue;
else
//i know this is wrong, but what is the right way of doing it?
cleanList[i][k] = dirtyList[i][j];
k++;
}
}
return cleanList;
}
修改
感谢您的所有评论，我现在已经开始工作了！与我的想法相反，正如Barmar指出的那样，没有任何问题：
cleanArray [i] [j] = dirtyArray [i] [k]
我的代码无效，因为我犯了很多其他错误，例如： -casting calloc的返回值 - 不为calloc正确分配内存 - 不正确的括号
我还在头文件中有代码，我认为它包含了自己的问题。
答案 0 :(得分：3)
假设您使用的是calloc size参数错误。其中一个char** cleanList = (char**)calloc(arrayLength, CHARLEN);或cleanList[i] = (char*)calloc(10,CHARLEN);是错误的。您也不应该转换malloc() / calloc()的返回值。出于易读性和代码流的目的，我还替换了if语句。您还写了for (j=0; j<strlen(dirtyList[i]+1);j++)而不是for (j=0; j<strlen(dirtyList[i])+1;j++)，因为strlen()计算的字符串长度没有\0。这是代码几乎没有变化。
char** cleanStrings (char**dirtyList, int arrayLength)
{
int i, j, k;
char **cleanList = calloc(arrayLength,sizeof * cleanList);
for (i=0; i<arrayLength; i++)
{
k= 0;
cleanList[i] = calloc(10,sizeof * cleanList[i]);
for (j=0; j<strlen(dirtyList[i])+1;j++)
{
if ((dirtyList[i][j] != '[') && (dirtyList[i][j] != '#') && (dirtyList[i][j] != '-') ){
cleanList[i][k] = dirtyList[i][j];
k++;
}
}
}
return cleanList;
}
答案 1 :(得分：2)
您没有为CHARLEN分配足够的内存。我假设sizeof(char)是cleanList，这是1个字节。但是char*的元素是char **cleanList = calloc(arrayLength, sizeof(char *));
，它是4或8个字节，分配太小了。它应该是：
malloc使用calloc或sizeof (T)时的一般规则是，乘数始终为T，其中*是目标类型，其中少于char **。因此，如果您要分配给sizeof(char *)，那就是{
"name": "ionic-hello-world",
"author": "Ionic Framework",
"homepage": "http://ionicframework.com/",
"private": true,
"scripts": {
"clean": "ionic-app-scripts clean",
"build": "ionic-app-scripts build",
"ionic:build": "ionic-app-scripts build",
"ionic:serve": "ionic-app-scripts serve"
},
"dependencies": {
"@angular/animations": "4.0.0",
"@angular/common": "4.0.0",
"@angular/compiler": "4.0.0",
"@angular/compiler-cli": "4.0.0",
"@angular/core": "4.0.0",
"@angular/forms": "4.0.0",
"@angular/http": "4.0.0",
"@angular/platform-browser": "4.0.0",
"@angular/platform-browser-dynamic": "4.0.0",
"@angular/platform-server": "4.0.0",
"@ionic-native/call-number": "3.8.0",
"@ionic-native/core": "^3.10.3",
"@ionic-native/device": "3.8.0",
"@ionic-native/facebook": "^3.10.3",
"@ionic-native/geolocation": "3.8.0",
"@ionic-native/google-plus": "^3.10.3",
"@ionic-native/in-app-browser": "^3.8.0",
"@ionic-native/social-sharing": "3.8.0",
"@ionic-native/splash-screen": "^3.9.2",
"@ionic-native/status-bar": "^3.10.3",
"@ionic/cloud-angular": "^0.12.0",
"@ionic/storage": "^2.0.1",
"font-awesome": "4.7.0",
"ionic-angular": "3.0.1",
"ionicons": "3.0.0",
"jssha": "2.2.0",
"rxjs": "5.1.1",
"sw-toolbox": "3.4.0",
"zone.js": "^0.8.4"
},
"devDependencies": {
"@ionic/app-scripts": "1.3.0",
"@types/jssha": "0.0.29",
"@types/node": "7.0.13",
"typescript": "~2.2.1"
},
"cordovaPlugins": [
"cordova-plugin-whitelist",
"cordova-plugin-console",
"cordova-plugin-statusbar",
"cordova-plugin-device",
"cordova-plugin-splashscreen",
"ionic-plugin-keyboard"
],
"description": "App desc"
}
。