海湾合作委员会 4.4.3 c89
我正在尝试显示地址。基本上,我只是想证明我显示的地址是正确的。
我想显示每个指向 char 'device_gc', 'device_mg', 'device_cc' 的指针数组的地址
所以我在我的主要功能中显示它们。但是,在我的 display_list 函数中,我只是想证明我正在显示正确的地址。输出是一样的。
希望你明白?
非常感谢您的任何建议。
#include <stdio.h>
void display_list(char ***dev_list);
int main(void)
{
char *device_gc[] = {"GCDEV01", "GCDEV02", "GCDEV03", "GCDEV04", "GCDEV05", "GCDEV06", NULL};
char *device_mg[] = {"MGDEV01", "MGDEV02", "MGDEV03", "GCDEV05", NULL};
char *device_cc[] = {"CCDEV01", "CCDEV02", "CCDEV03", "CCDEV04", "CCDEV05", NULL};
char **device_list[] = {device_gc, device_mg, device_cc, NULL};
printf("device_gc [ %p ]\n", (void*)*device_gc);
printf("device_mg [ %p ]\n", (void*)*device_mg);
printf("device_cc [ %p ]\n", (void*)*device_cc);
display_list(device_list);
return 0;
}
void display_list(char ***dev_list)
{
while(**dev_list != NULL) {
printf("dev [ %p ]\n", (void*)**dev_list++);
}
}
期望的输出:
device_gc [ 0x80485e0 ]
device_mg [ 0x8048610 ]
device_cc [ 0x8048628 ]
dev [ 0x80485e0 ]
dev [ 0x8048610 ]
dev [ 0x8048628 ]
我得到的实际输出是不同的,有时会导致核心转储。这是为什么?
最佳答案
两个小调整。在打印 main()
中的值之前,您不应该取消引用 'device_gc' 等;你应该只在 display_list()
中使用一个取消引用:
#include <stdio.h>
void display_list(char ***dev_list);
int main(void)
{
char *device_gc[] = {"GCDEV01", "GCDEV02", "GCDEV03", "GCDEV04", "GCDEV05", "GCDEV06", NULL};
char *device_mg[] = {"MGDEV01", "MGDEV02", "MGDEV03", "GCDEV05", NULL};
char *device_cc[] = {"CCDEV01", "CCDEV02", "CCDEV03", "CCDEV04", "CCDEV05", NULL};
char **device_list[] = {device_gc, device_mg, device_cc, NULL};
printf("device_gc [ %p ]\n", (void*)device_gc);
printf("device_mg [ %p ]\n", (void*)device_mg);
printf("device_cc [ %p ]\n", (void*)device_cc);
display_list(device_list);
return 0;
}
void display_list(char ***dev_list)
{
while(*dev_list != NULL) {
printf("dev [ %p ]\n", (void*)*dev_list++);
}
}
关于c - 获取指向 char 指针的指针数组的地址,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/3614875/