char * pname =" Harry" [英] char* pname = "Harry"
问题描述
以下声明;
char * p;
p =（char *）malloc（20）;
p =" harry" ; //< -------------它完全有效吗？
免费（p）;
Statement given below;
char * p ;
p = (char*) malloc(20);
p = "harry" ; // <------------- Is it perfectly valid ?
free(p);
推荐答案
er****@gmail.com 写道：
下面的陈述;
char * p;
p =（char *）malloc（20）;
你不必演员。
p =" harry" ; //< -------------它完全有效吗？
有效，但它没有做你所期望的，它改变p指向
string literal" harry"而不是malloc返回的内存。
free（p）;
Statement given below;
char * p ;
p = (char*) malloc(20);
You don''t have to cast.
p = "harry" ; // <------------- Is it perfectly valid ?
Valid, but it doesn''t do what you expect, it changes p to point to the
string literal "harry" rather than the memory returned by malloc.
free(p);
糟糕，你试图释放哈利。
-
Ian Collins。
Bad, you are attempting to free "harry".
--
Ian Collins.
er **** @ gmail.com 写道：
以下声明;
char * p;
p =（char *）malloc（20）;
失去演员阵容，除非你正在使用*非常旧的实现
（C89之前）。
p =harry ; //< -------------它完全有效吗？
完全有效，但不正确，考虑到上下文。而不是
复制字符串harry的内容。在
p指向的记忆中，你已经分配了字符串文字harry的地址。到p，
导致你忘记你刚分配的内存，这是一个内存泄漏的内存。
试试
strcpy（p，harry）;
代替。别忘了#include< string.h>。
免费（p）;
Statement given below;
char * p ;
p = (char*) malloc(20);
Lose the cast, unless you''re working with a *very* old implementation
(pre-C89).
p = "harry" ; // <------------- Is it perfectly valid ?
Perfectly valid, but not correct, given the context. Instead of
copying the contents of the string "harry" to the memory pointed to by
p, you''ve assigned the address of the string literal "harry" to p,
causing you to lose track of the memory you just allocated, which is a
memory leak.
Try
strcpy(p, "harry");
instead. Don''t forget to #include <string.h>.
free(p);
这将试图释放string literalharry，而不是你之前分配的内存
。
This will attempt to free the string literal "harry", not the memory
you allocated earlier.
John Bode写道：
John Bode wrote:
er****@gmail.com 写道：
下面的陈述;
char * p;
p =（char *）malloc（20）;
Statement given below;
char * p ;
p = (char*) malloc(20);
失去演员，除非你正在使用*非常旧的实现
（在C89之前）。
Lose the cast, unless you''re working with a *very* old implementation
(pre-C89).
[snip]
或代码应该可以用于C和C ++。
/ Peter
[snip]
Or the code should be usable with both C and C++.
/Peter
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