如何声明指向char*数组的指针数组
问题是访问指针数组的每个地址时,会将我的数据合并到后续的数据中如何声明指向char*数组的指针数组,c,C,问题是访问指针数组的每个地址时,会将我的数据合并到后续的数据中 正如在备注中所说,数组不是以空字符结尾的,这就是为什么printf在它们之后继续 得到这个结果是因为编译器将数组一个接一个地放在内存中,最后由于其他原因出现了一个空字符 所以最小的变化是 unsigned char mydata1[] = {0x41,0x42,0x43, 0}; unsigned char mydata2[] = {0x44,0x45,0x46, 0}; unsigned char mydata3[] = {0x4
正如在备注中所说,数组不是以空字符结尾的,这就是为什么printf在它们之后继续 得到这个结果是因为编译器将数组一个接一个地放在内存中,最后由于其他原因出现了一个空字符 所以最小的变化是
unsigned char mydata1[] = {0x41,0x42,0x43, 0};
unsigned char mydata2[] = {0x44,0x45,0x46, 0};
unsigned char mydata3[] = {0x47,0x48,0x49,0x4A, 0}
但这样做的前提是编译器使用ASCII代码,这是不可读的,最好这样做
unsigned char mydata1[] = {'A', 'B', 'C', 0};
unsigned char mydata2[] = {'D', 'E', 'F', 0};
unsigned char mydata3[] = {'G','H','I','J', 0};
或者更简单:
unsigned char mydata1[] = "ABC";
unsigned char mydata2[] = "DEF";
unsigned char mydata3[] = "GHIJ";
除此之外
做fori=0;i<3;i++是危险的,因为如果将元素数修改为charPtr,则还需要修改for,一种方法是添加一个空指针以标记指针列表的末尾,并迭代到一个空指针,另一种方法是在i#include <stdio.h>
#include <string.h>
/* String are NULL terminated, you might have read comments in question */
unsigned char mydata1[] = {0x41, 0x42, 0x43, 0x00};
unsigned char mydata2[] = {0x44, 0x45, 0x46, 0x00};
unsigned char mydata3[] = {0x47, 0x48, 0x49, 0x4A, 0x00};
/* Here I have used a NULL terminated array so that we can traverse
* even if we do not know the length of array
*/
unsigned char *charPtr[] =
{
mydata1,
mydata2,
mydata3,
NULL
};
int main()
{
int i;
for (i = 0; charPtr[i] != NULL; i++) {
printf("%s\n", charPtr[i]);
}
return 0;
}