[英]how to numerically integrate a variable that is being calculate in the program as a pointer (using e.g. trapezoidal rule) in c language
[英]Pointer to pointer argument variable and how they work? e.g. function(int **ptr)
我正在编写代码,以了解将指针作为函数中的参数传递时会发生什么。
void ptrTest(int **arg_ptr);
int main() {
int some_var = 5;
int *ptr1;
ptr1 = &some_var;
printf("Address of some_var: %u\n\n", &some_var);
printf("Value stored in *ptr1 (references to some_var): %u\n", ptr1);
printf("Address of *ptr1 variable: %u\n\n", &ptr1);
ptrTest(ptr1);
}
void ptrTest(int **arg_ptr){
printf("Value stored in **arg_ptr (references to some_var): %u\n",arg_ptr);
}
结果如下:
Address of some_var: 3119323004
Value stored in *ptr1 (references to some_var): 3119323004
Address of *ptr1 variable: 3119322992
Value stored in **arg_ptr (references to some_var): 3119323004
令我惊讶的是arg_ptr引用了指向some_var地址的值。 我期望** arg_ptr指向* ptr并存储3119322992的值(引用* ptr的地址)。
当我测试指向函数外部的指针时,它的行为确实如此。 为什么指针作为参数的指针不同呢?
你能告诉我这里发生了什么吗?
首先是什么指针? C / C ++中的指针与其他类型如int,char等的变量一样。但是此变量的特殊之处在于,它与其他变量不同,它仅保存内存位置的地址。 同样,该内存位置也可能是指针变量或任何其他常规变量(int或char)。
现在什么是指针指针? 可以存储指针变量的地址的变量,并且该指针变量可能包含另一个变量的地址,例如:
int i = 10; //`i` is assign with a value 10 and `i` has its own address which we can get by `&i`;
int *ptr1 = &i;// now ptr1 is pointer to `i` means ptr1 is assign with the
//address of `i` hence if we dereference the address of *ptr1 we will get the value stored at that memory location
现在你的情况
void ptrTest(int **arg_ptr){
printf("Address store in of **arg_ptr: %u\n",arg_ptr);
}
所以在这里它会像下面一样工作
int **arg_ptr = ptr1;//Wrong, `ptr1` is assign to `arg_ptr`, which is wrong because `arg_ptr` is a pointer to pointer type
所以在这里您应该存储指针的地址,但是要存储变量int的地址,即i
。 因为i
地址已在语句int * ptr1 =&i;中分配。 到ptr1。 正确的分配是
arg_ptr = &ptr1; //address of a pointer not address of a int variable.
现在首先取消引用:
*arg_ptr; //value of a pointer ptr1 that means address of `i`
*(*arg_ptr); or **arg_ptr;// we further dereferenced the address of ptr1 here, which is value 10
现在您应该像下面这样调用函数:
ptrTest(&ptr1);// address of a pointer.
ptrTest
需要一个int **
类型的参数,但是您正在传递int*
。 您的编译器应该抱怨。 您需要将ptr
的地址传递给函数。
ptrTest(&ptr1);
除此之外,您应该使用%p
规范来打印地址。
printf("Address of some_var: %p\n\n", (void*)&some_var);
当我编译您的代码时,我会得到一长串错误:
"test.c", line 11: warning #2181-D: argument is incompatible with
corresponding format string conversion
printf("Address of some_var: %u\n\n", &some_var);
^
"test.c", line 12: warning #2181-D: argument is incompatible with
corresponding format string conversion
printf("Value stored in *ptr1 (references to some_var): %u\n", ptr1);
^
"test.c", line 13: warning #2181-D: argument is incompatible with
corresponding format string conversion
printf("Address of *ptr1 variable: %u\n\n", &ptr1);
^
"test.c", line 14: warning #2167-D: argument of type "int *" is incompatible
with parameter of type "int **"
ptrTest(ptr1);
^
"test.c", line 18: warning #2181-D: argument is incompatible with
corresponding format string conversion
printf("Value stored in **arg_ptr (references to some_var): %u\n",arg_ptr);
让我们重写这段代码,使其编译时不会出错,并且可能会更清晰一些:
#include <stdio.h>
void ptrTest(int **arg_ptr)
{
printf("Value stored in arg_ptr (points to ptr1): %#p\n", arg_ptr);
printf("Value pointed to by arg_ptr (i.e. *arg_ptr - should be same as ptr1): %#p\n", *arg_ptr);
printf("Value pointed to by *arg_ptr (i.e. **arg_ptr - should be same as some_var): %d\n", **arg_ptr);
}
int main()
{
int some_var = 5;
int *ptr1;
ptr1 = &some_var;
printf("some_var: %d\n", some_var);
printf("Address of some_var: %#p\n\n", &some_var);
printf("Value stored in ptr1 (should be address of some_var): %#p\n", ptr1);
printf("Address of ptr1 variable: %#p\n\n", &ptr1);
ptrTest(&ptr1);
}
发生了什么变化:
在printf( %d
)中使用带符号的int格式,而不是无符号的格式( %u
)。
打印指针时,在printf( %p
)中使用的指针格式。
向ptrTest
添加了代码,以跟随arg_ptr
一直回到指针链的基本目标。
添加了代码以打印some_var
的值
更改了输出的用语,以澄清显示的内容以及应该匹配的值。
新版本可以正确编译(HP-UX默认C编译器)。
运行新版本时,将输出以下输出:
some_var: 5
Address of some_var: 0x7fffecd0
Value stored in ptr1 (should be address of some_var): 0x7fffecd0
Address of ptr1 variable: 0x7fffecd4
Value stored in arg_ptr (points to ptr1): 0x7fffecd4
Value pointed to by arg_ptr (i.e. *arg_ptr - should be same as ptr1): 0x7fffecd0
Value pointed to by *arg_ptr (i.e. **arg_ptr - should be same as some_var): 5
现在,您可以沿着指针链前后移动,以查看哪个指针指向哪个值,以及它们如何链接在一起。
祝你好运。
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