I need to allocate memory dynamically for an array of pointers.
我需要为指针数组动态分配内存。
Let us assume,
我们假设,
char *names[50];
char *element;
I used the following code to allocate memory dynamically which is producing error.
我使用以下代码动态分配内存,产生错误。
names=malloc(sizeof(char *));
Afterwards, i need to assign another character pointer to this one, say
然后,我需要为这个指定另一个字符指针,比方说
names=element;
I am getting error as ": warning: assignment from incompatible pointer type"
.
我收到错误为“:警告:从不兼容的指针类型分配”。
How can i resolve this?
我怎么解决这个问题?
3 个解决方案
#1
2
names=malloc(sizeof(char *));
names = malloc(sizeof(char *));
Will allocate either 4 or 8 bytes (depending on your system). This doesn't make sense since your array was already sized at 50 entries in the declaration...
将分配4或8个字节(取决于您的系统)。这没有意义,因为您的数组已在声明中的50个条目中调整大小...
names=element;
名称=元件;
This is not how arrays are used in C. You have declared that there are 50 elements in "names" and each one is allocated as a different pointer to an array of characters. You need to decide which element in the array you want to assign. For eaxample:
这不是C中如何使用数组。您已声明“names”中有50个元素,并且每个元素都被分配为指向字符数组的不同指针。您需要确定要分配的数组中的哪个元素。例如:
char *test1 = "test string 1";
char *test2 = "test string 2";
names[0] = test1; // Assign pointer to string 1 to first element of array
names[1] = test2; // Assign pointer to string 2 to second element of array
#2
1
If you want to dynamically allocate an array of N
char *
pointers, then you would use:
如果要动态分配N个char *指针的数组,那么您将使用:
char **names;
names = malloc(N * sizeof p[0]);
To assign the char *
value element
to the first element in the array, you would then use:
要将char * value元素分配给数组中的第一个元素,您将使用:
names[0] = element;
#3
1
You may want to check this tutorial out: http://dystopiancode.blogspot.com/2011/10/dynamic-multidimensional-arrays-in-c.html It says how to allocate dynamic memory for arrays,matrices,cubes and hypercube and also how you can free it.
您可能需要查看本教程:http://dystopiancode.blogspot.com/2011/10/dynamic-multidimensional-arrays-in-c.html它说明如何为数组,矩阵,多维数据集和超立方体分配动态内存你也可以如何释放它。
#1
2
names=malloc(sizeof(char *));
names = malloc(sizeof(char *));
Will allocate either 4 or 8 bytes (depending on your system). This doesn't make sense since your array was already sized at 50 entries in the declaration...
将分配4或8个字节(取决于您的系统)。这没有意义,因为您的数组已在声明中的50个条目中调整大小...
names=element;
名称=元件;
This is not how arrays are used in C. You have declared that there are 50 elements in "names" and each one is allocated as a different pointer to an array of characters. You need to decide which element in the array you want to assign. For eaxample:
这不是C中如何使用数组。您已声明“names”中有50个元素,并且每个元素都被分配为指向字符数组的不同指针。您需要确定要分配的数组中的哪个元素。例如:
char *test1 = "test string 1";
char *test2 = "test string 2";
names[0] = test1; // Assign pointer to string 1 to first element of array
names[1] = test2; // Assign pointer to string 2 to second element of array
#2
1
If you want to dynamically allocate an array of N
char *
pointers, then you would use:
如果要动态分配N个char *指针的数组,那么您将使用:
char **names;
names = malloc(N * sizeof p[0]);
To assign the char *
value element
to the first element in the array, you would then use:
要将char * value元素分配给数组中的第一个元素,您将使用:
names[0] = element;
#3
1
You may want to check this tutorial out: http://dystopiancode.blogspot.com/2011/10/dynamic-multidimensional-arrays-in-c.html It says how to allocate dynamic memory for arrays,matrices,cubes and hypercube and also how you can free it.
您可能需要查看本教程:http://dystopiancode.blogspot.com/2011/10/dynamic-multidimensional-arrays-in-c.html它说明如何为数组,矩阵,多维数据集和超立方体分配动态内存你也可以如何释放它。