f****4
发帖数: 1359 | 1 Explain the difference between the following two functions, highlight any
perceived problems, and state which one is the preferred implementation and
why it is preferred.
char *function1(void)
{
char *s = "12345";
return s;
}
char *function2(void)
{
char s[MAX_STRLEN] = "12345";
return s;
}
这有啥区别啊？ |
y*c
发帖数: 904 | 2 1. returned a pointer to a literal string, which is in the static memory
zone. Safe. But it should return const char*
2. returned a pointer to a local memory, which will be released after the
call. Dangerous. |
f*******r
发帖数: 1086 | 3 Great answer:)
const char *function1(void)
{
const char *s = "12345";
return s;
}
the function1 should be like this.
【在 y*c 的大作中提到】
: 1. returned a pointer to a literal string, which is in the static memory
: zone. Safe. But it should return const char*
: 2. returned a pointer to a local memory, which will be released after the
: call. Dangerous.
|
y**i
发帖数: 1112 | 4 问个问题，第二个函数里面的文字常量是不是也会在静态区里创建内存，然后再拷贝到
栈里的数组，还是根本就不存在这样一个文字常量？还有就是这个文字常量的作用域怎
样？
【在 y*c 的大作中提到】
: 1. returned a pointer to a literal string, which is in the static memory
: zone. Safe. But it should return const char*
: 2. returned a pointer to a local memory, which will be released after the
: call. Dangerous.
|
r****o
发帖数: 1950 | 5 刚才试了一下，两个const按存在不存在取4种组合，函数都能正确工作，除了一些
warning以外。
const char *function1(void)
{
const char *s = "12345";
return s;
}
const char *function2(void)
{
char *s = "12345";
return s;
}
char *function1(void)
{
const char *s = "12345";
return s;
}
char *function1(void)
{
char *s = "12345";
return s;
}
是不是这4个函数等价?
【在 f*******r 的大作中提到】
: Great answer:)
: const char *function1(void)
: {
: const char *s = "12345";
: return s;
: }
: the function1 should be like this.
|
y*c
发帖数: 904 | 6
I think you are right. It exists. I guess it's globally accessible. So if
"12345" is accessed at another place. It should have the same address. I
am not 100% sure about this.
【在 y**i 的大作中提到】
: 问个问题，第二个函数里面的文字常量是不是也会在静态区里创建内存，然后再拷贝到
: 栈里的数组，还是根本就不存在这样一个文字常量？还有就是这个文字常量的作用域怎
: 样？
|
f*******r
发帖数: 1086 | 7 我觉得不是，应该是test的函数本身很简单，
stack空间没有被occupy，返回一个local的指针变量
应该是不safe的
【在 r****o 的大作中提到】
: 刚才试了一下，两个const按存在不存在取4种组合，函数都能正确工作，除了一些
: warning以外。
: const char *function1(void)
: {
: const char *s = "12345";
: return s;
: }
: const char *function2(void)
: {
: char *s = "12345";
|
f*******r
发帖数: 1086 | 8 恩，其实看VC的asm code应该就知道了:)
【在 y*c 的大作中提到】
:
: I think you are right. It exists. I guess it's globally accessible. So if
: "12345" is accessed at another place. It should have the same address. I
: am not 100% sure about this.
|
y**i
发帖数: 1112 | 9 你有没有尝试第三和第四种情况能不能修改字符元素呢？我觉得应该不能，尽管返回值
是char*
【在 r****o 的大作中提到】
: 刚才试了一下，两个const按存在不存在取4种组合，函数都能正确工作，除了一些
: warning以外。
: const char *function1(void)
: {
: const char *s = "12345";
: return s;
: }
: const char *function2(void)
: {
: char *s = "12345";
|
