[英]Returning a pointer to an array of ints in c++ (converted from string) - “invalid conversion” error
Header: 标头:
#ifndef HEADER_H_INCLUDED
#define HEADER_H_INCLUDED
int i[8];
#endif
Main: 主要:
#include <iostream>
#include <string>
#include "header.h"
using namespace std
int main(){
int *test;
string bits = "10011011";
*test = func(bits); // ERROR 2 HERE
//my goal here is to have a pointer in main that
//points to the 1 address of the global var array i
} //if im totally missing the point and there is a
//better way to do this please let me know
Function: 功能:
#include <iostream>
#include <string>
using namespace std
int *func(string str){
int l = str.size();
int *ptr;
for(int k=0; k < l; ++k){
i[k] = s[k] - '0';
}
*ptr = i; // ERROR 1 HERE
return ptr;
}
Hi all, 大家好,
When I attempt to compile the above code, I get two errors as labeled. 当我尝试编译以上代码时,出现两个错误,如标签所示。 They are:
他们是:
Error 1: invalid conversiopn from int* to int Error 2: invalid conversion from int to *int 错误1:从int *到int的无效对话错误2:从int到* int的无效转换
It seems I must have a fundamental understanding of what I am working with here. 似乎我必须对在这里使用的内容有基本的了解。 Am I not setting *ptr to point at the first memory address of the array i[]?
我不是将* ptr设置为指向数组i []的第一个内存地址吗? Why is the compiler telling me that I am trying to set the value of the pointer to the value of i[]?
为什么编译器告诉我我正在尝试将指针的值设置为i []的值? I want to set the value of the pointer to the ADDRESS of i[].
我想将指针的值设置为i []的ADDRESS。 If I add an & I get the same error.
如果我添加&,则会出现相同的错误。
Error 2 obviously follows error 1. It is the same error, just backwards. 错误2显然跟随错误1。这是相同的错误,只是向后。
My question is, then, what the heck am I doing wrong when I am trying to point *ptr in func() at the array i[]? 我的问题是,当我试图将func()中的* ptr指向数组i []时,我到底在做什么错? I DID try finding the answers elsewhere to no avail.
我想找到其他地方的答案都无济于事。
Thanks. 谢谢。
edit: this code is a rough transcription of the original (from another PC), so if you try to compile it and there are typos I apologize. 编辑:这段代码是原始代码(从另一台PC上)的粗略复制,因此,如果您尝试对其进行编译,并且有错别字,我深表歉意。
int *ptr;
declares a pointer to int
, so ptr
is a pointer: a variable that stores the memory location of some int
. 声明一个指向
int
的指针,所以ptr
是一个指针:一个存储某个int
的内存位置的变量。 Currently it doesn't point to anything in particular, just some random place in memory that you probably don't own. 当前,它并没有特别指向任何内容,只是您可能不拥有的内存中的随机位置。
*ptr
dereferences ptr
, so *ptr
is the int
that ptr
points to. *ptr
取消引用ptr
,因此*ptr
是ptr
指向的int
。
When you do *ptr = i;
当你做
*ptr = i;
you are attempting to set the int
pointed to by ptr
equal to i
, which is an array of 8 int
s. 您正在尝试将
ptr
指向的int
设置为i
,它是8 int
的数组。 In this context, the array decays into a pointer to its first element, so you're attempting to set an int
equal to an int*
, hence the error. 在这种情况下,数组会衰减为指向其第一个元素的指针,因此您尝试将
int
设置为等于int*
,从而导致错误。
This should just be ptr = i;
这应该只是
ptr = i;
, but even that isn't necessary, since you could just return i;
,但这不是必需的,因为您可以
return i;
and i
would decay into a pointer to the first element of the array. 而且
i
会变成一个指向数组第一个元素的指针。
Similarly, *test = func(bits);
同样,
*test = func(bits);
attempts to set the int
pointed to by test
equal to the pointer returned by func
. 尝试将
test
指向的int
设置为等于func
返回的指针。 It should just be test = func(bits);
它应该只是
test = func(bits);
, in which case the address that test
contains will become the one that func
returned. ,在这种情况下,
test
包含的地址将成为func
返回的地址。 In this case test
will point to the first element of i
. 在这种情况下,
test
将指向i
的第一个元素。
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