为什么malloc（）会修改此输出的行为？ [英] Why would malloc() modify the behavior of this output?
问题描述
如果符合以下条件：
#include< stdlib.h>
void * test_me（int a）{
int * m;
m =& a;
}
main（void）{
char * j;
j = test_me（1）;
printf（" j的值）是：％x，* j）;
}
我得到：
j的值是：1
然而，当我添加malloc（）
#include< stdlib.h>
void * test_me（int a）{
int * m;
m =& a;
}
main（无效）{
char * p;
char * j;
p = malloc（10 ）;
j = test_me（1）;
printf（" p的值是：％x，* p）;
printf（" j的值是：％x，* j）;
免费（p）;
}
我得到以下内容：
p的值为：0
j的值为：ffffffc4
为什么p变为零？
If go like the following:
#include <stdlib.h>
void *test_me(int a) {
int *m;
m = &a;
}
main(void){
char *j;
j = test_me(1);
printf("The value of j is : %x", *j);
}
I get:
The value of j is : 1
However, when I add malloc()
#include <stdlib.h>
void *test_me(int a) {
int *m;
m = &a;
}
main(void){
char *p;
char *j;
p = malloc(10);
j = test_me(1);
printf("The value of p is : %x", *p);
printf("The value of j is : %x", *j);
free(p);
}
I get the following:
The value of p is : 0
The value of j is: ffffffc4
Why does p become zero?
推荐答案
grocery_stocker写道：
grocery_stocker wrote:
如果如下所示：
#include< stdlib.h>
void * test_me（int a）{
int * m;
m =& a;
请注意，您没有从此功能返回任何内容，
虽然其签名表示将返回指向void的指针。
未定义的行为。所有投注都已关闭。
}
main（无效）{
char * j;
j = test_me（1）;
再次，由于test_me（）无法返回任何内容，因此`j''的值为
不确定。
printf（& j的值是：％x，* j）;
}
我得到：
j的值是：1
但是，当我添加malloc（）时
#include< stdlib.h>
void * test_me（int a）{
int * m;
m =& a;
}
main（无效）{
char * p;
char * j;
p = malloc（10）;
j = test_me（1）;
printf（" p的值是：％x，* p）;
printf（" j的值是：％x，* j）;
免费（p）;
}
我得到以下内容： br /> p的值是：0
j的值是：ffffffc4
为什么p变为零？
If go like the following:
#include <stdlib.h>
void *test_me(int a) {
int *m;
m = &a;
Please notice that you''re not returning anything from this function,
though its signature indicates a pointer to void will be returned.
Undefined behavior. All bets are off.
}
main(void){
char *j;
j = test_me(1);
Again, since test_me() fails to return anything, the value of `j'' is
indeterminate.
printf("The value of j is : %x", *j);
}
I get:
The value of j is : 1
However, when I add malloc()
#include <stdlib.h>
void *test_me(int a) {
int *m;
m = &a;
}
main(void){
char *p;
char *j;
p = malloc(10);
j = test_me(1);
printf("The value of p is : %x", *p);
printf("The value of j is : %x", *j);
free(p);
}
I get the following:
The value of p is : 0
The value of j is: ffffffc4
Why does p become zero?
未定义的行为意味着结果可能是*任何东西* - 甚至鼻子
恶魔可能随之而来。
HTH，
--ag
-
Artie Gold - 德克萨斯州奥斯汀
http://it-matters.blogspot.com （新帖子12/5）
http://www.cafepress.com/goldsays
Undefined behavior means the results could be *anything* -- even nasal
demons might ensue.
HTH,
--ag
--
Artie Gold -- Austin, Texas
http://it-matters.blogspot.com (new post 12/5)
http://www.cafepress.com/goldsays
grocery_stocker写道：
grocery_stocker wrote:
如果符合以下条件：
#include< stdlib.h>
void * test_me（int a）{
int * m;
m =& a;
}
main（void）{
char * j;
j = test_me（1）;
printf（" j的值为：％x"，* j）;
}
我得到：
j的值是：1
然而，当我添加malloc（）时
＃包括< stdlib.h>
void * test_me（int a）{
int * m;
m =& a;
}
main（void）{
char * p;
char * j;
p = malloc（10）;
j = test_me（1） ;
printf（" p的值是：％x，* p）;
printf（" value） j是：％x，* j）;
免费（p）;
}
我得到以下内容：
价值p是：0
j的值是：ffffffc4
为什么p变为零？
If go like the following:
#include <stdlib.h>
void *test_me(int a) {
int *m;
m = &a;
}
main(void){
char *j;
j = test_me(1);
printf("The value of j is : %x", *j);
}
I get:
The value of j is : 1
However, when I add malloc()
#include <stdlib.h>
void *test_me(int a) {
int *m;
m = &a;
}
main(void){
char *p;
char *j;
p = malloc(10);
j = test_me(1);
printf("The value of p is : %x", *p);
printf("The value of j is : %x", *j);
free(p);
}
I get the following:
The value of p is : 0
The value of j is: ffffffc4
Why does p become zero?
我的猜测是在调用test_me之后堆栈已损坏，
因为test_me返回未指定的值，因此p可能由
printf修改。
My guess is that after test_me has been called the stack is corrupt,
since test_me returns an unspecified value, so p is probably modified by
printf.
换句话说，在函数test_me（）中，我应该使用
''static''？
So in other words, when in the function test_me(), I should have used
''static''?
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