c/c++ 指针数组

发布于:2024-09-18 ⋅ 阅读:(125) ⋅ 点赞:(0)

        顾名思义就是元素为指针的数组,如定义一个 char *类型的指针数组:

 char *strName[] = {
	"123",
	"456",
	"789",
	"abc",
	"def"
};

如果是这样初始化的指针数组,c++编译是会抛出警告信息的:

因为用了字符串常量进行初始化,所以是不可被修改的,所以定义时需要加上 const:

#include <stdio.h>
#include <stdlib.h>

 char *strName[] = {
	"123",
	"456",
	"789",
	"abc",
	"def"
};

int main()
{
	printf("strName[1] = %s\n", strName[1]);
	printf("*strName = %s\n", *strName);
	printf("*strName+1 = %s\n", *strName+1);
	printf("sizeof strName[2] = %lu\n", sizeof(strName[2]));
	
    return 0;
}

*strName 是解引用,所以输出是 “123” ,而 *strName + 1 为什么输出是 “23” 呢?因为在 *strName 解引用后它的类型是 char*,+1 后即挪动一个 char 长度,指向了“123”中的“2”位置,其实它跟 “123” + 1 的效果是一样的:
 

#include <stdio.h>
#include <stdlib.h>

const char *strName[] = {
	"123",
	"456",
	"789",
	"abc",
	"def"
};

int main()
{
	printf("strName[1] = %s\n", strName[1]);
	printf("*strName = %s\n", *strName);
	printf("*strName+1 = %s\n", *strName+1);
	printf("\"123\"+1 = %s\n", "123"+1);
	printf("sizeof strName[2] = %lu\n", sizeof(strName[2]));
	
    return 0;
}

如果不用字符串常量初始化,也是可以手动申请内存的,但定义是就不能用const了,如:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char *strName[3];
int main()
{
	char buf[16];
	for(int i = 0; i < 3; i++)
	{
		strName[i] = new char[16];
		snprintf(buf, sizeof(buf), "test_%d", i * 10);
		strncpy(strName[i], buf, 15);
	}
	printf("strName[1] = %s\n", strName[1]);
	printf("*strName = %s\n", *strName);
	printf("*strName+1 = %s\n", *strName+1);
	printf("sizeof strName[2] = %lu\n", sizeof(strName[2]));
	
	for(int i = 0; i < 3; i++)
	{
		if(strName[i])
		{
			delete strName[i];
		}
	}
    return 0;
}

 


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