C语言 c - 如何在c中将char *转换为char []
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how to convert from char* to char[] in c
提问by Bassel Shawi
here is a code sample
这是一个代码示例
void something()
{
char c[100];
scanf("%s",c);
char c2[100]=c;
}
my problem is when i do this assignment an error says that i can't assign
我的问题是当我做这个任务时，错误说我不能分配
char * "c" to char[] "c2";
how can i achieve this assignment?
我怎样才能完成这项任务？
回答by John Bode
You'll have to use strcpy()(or similar):
您必须使用strcpy()（或类似的）：
...
char c2[100];
strcpy(c2, c);
You can't assign arrays using the =operator.
您不能使用=运算符分配数组。
回答by Fred Larson
You need to use strcpy()
你需要使用 strcpy()
char c2[100];
strcpy(c2, c);
回答by aiGuru
Better practice would be to use strncpy(c2, c, 100) to avoid buffer overflow, and of course limit the data entry too with something like scanf("%99s", c);
更好的做法是使用 strncpy(c2, c, 100) 来避免缓冲区溢出，当然也可以使用 scanf("%99s", c); 之类的方法限制数据输入。
回答by Chris Dodd
char []is not a valid value type in C (its only a valid declaration type), so you can't actualy do anything with char []types. All you can do is convert them to something else (usually char *) and do something with that.
char []在 C 中不是一个有效的值类型（它只是一个有效的声明类型），所以你实际上不能对char []类型做任何事情。您所能做的就是将它们转换为其他东西（通常char *）并对其进行处理。
So if you wany to actually do something with the data in the array, you need to use some function or operation that takes a char *and derefences it. Obvious choices for your example are strcpyor memcpy
因此，如果您想对数组中的数据进行实际操作，则需要使用一些函数或操作来获取 achar *并取消对它的引用。您的示例的明显选择是strcpy或memcpy
