寻找malloc（）的帮助 [英] Looking for malloc() help
问题描述
我正在学习C并且有一个问题：malloc（）。
我写了一个简单的程序，它为一个结构赋值然后
打印如下：
#include< stdio.h>
#include< stdlib.h>
struct item {
char name [20];
int quantity;
};
int main（int argc，char * argv []）
{
struct item * stuff;
/ /为结构分配内存
stuff = malloc（3 * sizeof（struct item））;
strcpy（stuff [1] .name，" apple" ;）;
stuff [1] .quantity = 1;
strcpy（stuff [2] .name，" banana"）;
stuff [2] .quantity = 2;
printf（"％s％d \ n"，stuff [1] .name，st.quantity）;
printf（"％s％d \ n"，stuff [2] .name，st.quantity）;
免费（东西）;
返回0;
}
然后我改变结构d eclaration以动态
为char数组分配内存
：
struct item {
char * name ;
int数量;
};
程序编译没有错误，但它崩溃了以下
错误：
第3行：2485分段错误
感谢您的帮助
我不知道是什么问题。以下代码适用于我的
机器。
struct item {
char * name;
int quantity;
};
int main（）
{
struct item * stuff;
//为结构分配内存
stuff =（struct item *）malloc（3 * sizeof（struct item））;
stuff [1] .name =（char *）malloc（sizeof（" apple"）））;
strcpy（stuff [1] .name，" apple"）;
stuff [1] .quantity = 1;
stuff [2] .name =（char *）malloc（sizeof（" banana））;
strcpy（stuff [2] .name，" banana"）;
stuff [2] .quantity = 2;
printf（"％s ％d \ n"，stuff [1] .name，stuff [1] .quantity）;
printf（"％s％d \ n"，stuff [2] .name，东西[2] .quantity）;
免费（东西）;
返回0;
}
xiaohuamao写道：
我不知道是什么问题。以下代码适用于我的
机器。
struct item {
char * name;
int quantity;
};
int main（）
{
struct item * stuff;
//为结构分配内存
stuff =（struct item *）malloc（3 * sizeof（struct item））;
stuff [1] .name =（char *）malloc（sizeof（" apple"）））;
strcpy（stuff [1] .name，" apple"）;
stuff [1] .quantity = 1;
stuff [2] .name =（char *）malloc（sizeof（" banana））;
strcpy（stuff [2] .name，" banana"）;
stuff [2] .quantity = 2;
printf（ "％s％d \ n"，stuff [1] .name，stuff [1] .quantity）;
printf（"％s％d \ n"，stuff [2 ] .name，stuff [2] .quantity）;
免费（东西）;
返回0;
}
不确定是否重要，我的电脑如下：
Linux Slackware 10.1
使用gcc -o test struct.c编译程序
还有其他人能够运行吗？
SP写道：
我正在学习C并且有一个问题re：malloc（）。
我写了一个简单的程序，它为一个结构赋值然后
将其打印如下：
#include< stdio.h>
#include< stdlib.h> ;
struct item {
char name [20];
int quantity;
} ;
int main（int argc，char * argv []）
{
struct item * stuff;
//为结构分配内存
stuff = malloc（3 * sizeof（struct item））;
strcpy（ stuff [1] .name，" apple"）;
stuff [1] .quantity = 1;
strcpy（stuff [2] .name，" ban ana"）;
stuff [2] .quantity = 2;
printf（"％s％d \ n"，stuff [1] .name，st.quantity）;
st尚未定义。你可能意味着东西[1] .quantity
printf（"％s％d \ n"，stuff [2] .name，st.quantity）;
stuff [2] .quantity
>
免费（东西）;
返回0;
}
然后我改变结构声明以便动态地分配
分配内存
为char数组：
struct item {
char * name;
int quantity;
};
程序编译时没有错误，但它崩溃了以下
错误：
在第一种情况下，struct包含用于存储
名称的内存（20字节），因此结构的malloc（）就足够了。在第二个
的情况下，struct只包含一个指针，因此你要复制apple。
无论随机地址最终都是[1] .name。未定义的行为。[*]
您可以使用malloc（）分配空间来存储名称
并将结果指针分配给stuff [1] .name，或者只是做
stuff [1] .name =" apple" ;;
并且根本不复制字符串。
在这种情况下，只需指定指针就会更有意义，但在
实例中，任何一种策略都可能是合适的。
-thomas
[*]从技术上讲，未定义的行为可能比复制到某个随机地址更糟糕，但这已经够糟了。
I am learning C and have a question re: malloc().
I wrote simple program which assigns a value to a structure and then
prints it as follow:
#include <stdio.h>
#include <stdlib.h>
struct item {
char name[20];
int quantity;
};
int main (int argc, char *argv[])
{
struct item *stuff;
//allocate memory for structure
stuff = malloc(3 * sizeof(struct item));
strcpy(stuff[1].name, "apple");
stuff[1].quantity = 1;
strcpy(stuff[2].name, "banana");
stuff[2].quantity = 2;
printf("%s %d\n", stuff[1].name, st.quantity);
printf("%s %d\n", stuff[2].name, st.quantity);
free(stuff);
return 0;
}
I then change the structure declaration in order to dynamically
allocate memory
for the char array:
struct item {
char *name;
int quantity;
};
The program compiles with no errors, but it crashes with the following
error:
line 3: 2485 Segmentation fault
Thanks for your help
I don''t know what is the problem. The following code works in my
machine.
struct item {
char* name;
int quantity;
};
int main ()
{
struct item *stuff;
//allocate memory for structure
stuff = (struct item *)malloc(3 * sizeof(struct item));
stuff[1].name = (char *) malloc(sizeof("apple"));
strcpy(stuff[1].name, "apple");
stuff[1].quantity = 1;
stuff[2].name = (char *) malloc(sizeof("banana"));
strcpy(stuff[2].name, "banana");
stuff[2].quantity = 2;
printf("%s %d\n", stuff[1].name, stuff[1].quantity);
printf("%s %d\n", stuff[2].name, stuff[2].quantity);
free(stuff);
return 0;
}
xiaohuamao wrote:I don''t know what is the problem. The following code works in my
machine.
struct item {
char* name;
int quantity;
};
int main ()
{
struct item *stuff;
//allocate memory for structure
stuff = (struct item *)malloc(3 * sizeof(struct item));
stuff[1].name = (char *) malloc(sizeof("apple"));
strcpy(stuff[1].name, "apple");
stuff[1].quantity = 1;
stuff[2].name = (char *) malloc(sizeof("banana"));
strcpy(stuff[2].name, "banana");
stuff[2].quantity = 2;
printf("%s %d\n", stuff[1].name, stuff[1].quantity);
printf("%s %d\n", stuff[2].name, stuff[2].quantity);
free(stuff);
return 0;
}Not sure it matters, my PC is as follows:
Linux Slackware 10.1
compiling program with "gcc -o test struct.c"
Is anyone else able to run this ?
SP wrote:I am learning C and have a question re: malloc().
I wrote simple program which assigns a value to a structure and then
prints it as follow:
#include <stdio.h>
#include <stdlib.h>
struct item {
char name[20];
int quantity;
};
int main (int argc, char *argv[])
{
struct item *stuff;
//allocate memory for structure
stuff = malloc(3 * sizeof(struct item));
strcpy(stuff[1].name, "apple");
stuff[1].quantity = 1;
strcpy(stuff[2].name, "banana");
stuff[2].quantity = 2;
printf("%s %d\n", stuff[1].name, st.quantity);st has not been defined. You probably mean stuff[1].quantity
printf("%s %d\n", stuff[2].name, st.quantity);stuff[2].quantity
>
free(stuff);
return 0;
}
I then change the structure declaration in order to dynamically
allocate memory
for the char array:
struct item {
char *name;
int quantity;
};
The program compiles with no errors, but it crashes with the following
error:In the first case the struct contains memory (20 bytes) for storing the
names so the malloc() for the structs is sufficient. In the second
case the struct contains only a pointer, so you are copying "apple" to
whatever random address ends up in stuff[1].name. Undefined behaviour.[*]
You could either use malloc() to allocate space for storing the name
and assign the resulting pointer to stuff[1].name, or just do
stuff[1].name = "apple";
and not copy the string at all.
In this case just assigning the pointer would make more sense, but in a
real example either strategy might be appropriate.
-thomas
[*] Technically the undefined behaviour could be even worse than
copying to some random address, but that''s bad enough.
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