无效的操作数错误 [英] invalid operand error
问题描述
我很确定这是算术中的一个简单错误，但我无法绕过它......我不断得到这个错误：无效操作数到二进制+（有int *和int *）。
void sort_ver2（ int *指针， int size）
{
int * i，* j;
int temp;
for （i =指针; i<（指针+大小）; ++ i）{
for （j =指针+ 1 ; j<（指针+大小）; ++ j）{
if （（pointer + i）<（pointer + j））{
temp =（pointer + j）;
（指针+ j）=（指针+ i）;
（指针+ i）= temp;
}
}
}
}
提前致谢！
仔细看看：
for （i =指针; i < ;（指针+大小）; ++ i）{
^
这应该在这里吗？或者你的意思是=？
抱歉:)更新了帖子，我的意思是<冒号不应该在那里。
看看新版本，我是不完全确定你要做什么，但我能理解为什么编译器在抱怨：
int * i，* j;
int temp;
...
temp =（指针+ j）;
指针和j都是指针到整数的变量 - 和你不能添加两个指针，因为没有什么明智的你得到！这有点像尝试添加东方和左 - 它不会给你带来任何有用的结果。
因为这是一个排序函数，我怀疑你最好用i和j写作int而不是int *并使用它们添加到基础指针值。
并且还使用*（指针+ i）来访问值可能会有所帮助！ ：笑：
我相信的问题似乎源于你对指针的理解和使用。
你的代码指针+ i真的没有任何意义。我将尝试用简单的图表来说明。
想象一下，你有一个包含一些数据的数组。我们假设*指针保存第一个元素的地址。我们还假设内存中的任意位置为0x4003000。
- --------------------
| a | b | c | d | \0 |
- --------------------
所以，
'a'是0x4003000，
'b'位于0x4003001，
'c'位于0x4003002，
'd'位于0x4003003
和
NULL终止符位于0x4003004
所以，现在我们已经有了这个，让我们从你的第一个for循环中包含你的第3行 - 假设大小为4.
for （i =指针; i < （指针+大小）; ++ i）{
for （j =指针+ 1 ; j < （指针+大小）; ++ j）{
if （（指针+ i）< （指针+ j））{
这将变为
< pre lang =c ++> for （i = 0x4003000; I< 0x4003000 + 4; ++ i）
{
for （j = 0x4003001; j< 0x4003000 + 4; ++ j）
{
if （（0x4003000 + 0x4003000）<（0x4003000 + 0x4003001））
开始看到问题了吗？！
解决方案是，我认为:(你的意图不是1000％清楚）
void sort_ver2（ int *指针， int size）
{
int * i，* j;
int temp;
for （i =指针; i<（指针+大小）; ++ i）
{
for （j =指针+ 1 ; j<（指针+大小）; ++ j）
{
if （* i< * j）
{
temp = * j;
* j = * i;
* i = temp;
}
}
}
}
我假设你想要比较i和j指向的_values_，如果满足条件则交换它们？
一种真正可怕的（但是完全有效和合法的）使用指针的方式，在我的拙见中。 ：）
原因是C ++不允许添加指针。
你对增加指针有什么期望？指针是内存地址。添加2个指针将指向（可能）整个不同位置的另一个内存地址。由于这是没用的，C ++禁止指针添加。
我认为你对将指针添加到指针感到困惑。以下代码在C ++中有效，并且有意义。
int arr [ 5 ] = { 1 ， 2 ， 3 ， 4 ， 5 };
int * p =& arr;
int * p_third =（p + 3 ）;< br $>
arr是一个包含5个整数的数组。现在，p指向数组的第一个元素（arr [0]）。并且p_third指针被赋值为（p + 3）。什么是（p + 3）？
在C ++中，上面代码的最后一行也可以写得像这样：
int * p_third =（p + 3 * sizeof （ int ））;
所以，（p + 3）实际上是（p + 3 * sizeof（int ））。
最后，p_third指向arr [3]。
提示：你应该像Java人一样停止编码。像这样的代码（它更具可读性和全面性）：
if （）
{
if （）
{
if （）
{
}
}
else
{
}
}
I'm pretty sure it's a simple error in the arithmetic but I just cant wrap my head around it..I keep getting this error: invalid operand to binary +(have int* and int*).
void sort_ver2(int *pointer, int size)
{
int *i, *j;
int temp;
for(i = pointer; i < (pointer+size); ++i){
for(j = pointer + 1; j < (pointer+size); ++j){
if((pointer + i) < (pointer + j)){
temp=(pointer + j);
(pointer + j)=(pointer + i);
(pointer + i)=temp;
}
}
}
}
Thanks in advance!
Have a close look:
for(i = pointer; i <; (pointer+size); ++i){ ^Should this be here? Or did you mean "="?
"Sorry :) updated the post, what i meant was simply "<" the colon isnt't supposed to be there."
Looking at the new version, I'm not exactly sure what you are trying to do, but I can understand why the compiler is complaining:
int *i, *j; int temp; ... temp=(pointer + j);Both
pointerandjare pointer-to-integer variables - and you can't add two pointers, because there is nothing sensible you get! It's a bit like trying to add "East" and "Left" - it doesn't leave you anything useful as a result.
Since this is a sort function, I would suspect you would be better off writing it withiandjasintrather thanint*and using them to add to your basepointervalue.
And also using *(pointer+i) to access the values would probably help! :laugh:
The problem I believe, appears to stem from your understanding and use of pointers.
Your codepointer + ireally doesn't make any sense. I'll try to illustrate with a simple diagram.
Imagine you had an array with some data in it. We'll assume that *pointer holds the address of the first element. We'll also assume an arbitrary location in memory of 0x4003000.
---------------------- | a | b | c | d | \0 | ----------------------
So,
'a' is at 0x4003000,
'b' is at 0x4003001,
'c' is at 0x4003002,
'd' is at 0x4003003
and the
NULL terminator is at 0x4003004
So, now we have that in place, lets run through the 3 lines from and including your first for loop - assuming size to be 4.
for(i = pointer; i < (pointer+size); ++i){ for(j = pointer + 1; j < (pointer+size); ++j){ if((pointer + i) < (pointer + j)){
This will turn into
for (i=0x4003000; i<0x4003000+4; ++i) { for (j=0x4003001; j<0x4003000+4; ++j) { if ((0x4003000+0x4003000) < (0x4003000+0x4003001))
Starting to see the problem yet?!
The solution is, I think: (your intention isn't 1000% clear to me)
void sort_ver2(int *pointer, int size) { int *i, *j; int temp; for(i = pointer; i < (pointer+size); ++i) { for(j = pointer + 1; j < (pointer+size); ++j) { if( *i < *j) { temp = *j; *j = *i; *i = temp; } } } }
I assume you'd like to compare the _values_ that i and j point to, swapping them if a condition is met?
A truly horrid (but perfectly valid & legal) way to use pointers, in my humble opinion. :)
The reason is C++ does not allow pointer addition.
What do you expect by adding pointers ? A pointer is a memory address. Adding 2 pointers will point to (maybe) another memory address in an entire different location. Since this is no use, C++ prohibits pointer addition.
I think you are confused about adding integers to pointers. The following code is valid in C++, and makes sense.
int arr[5] = {1, 2, 3, 4, 5}; int* p = &arr; int* p_third = (p + 3);
arris an array of 5 integers. Now,ppoints to the array's first element(arr[0]). Andp_thirdpointer is assigned to a value of(p + 3). So what's(p+3)?
In C++, the last line of the above code can be also written like this:
int* p_third = (p + 3 * sizeof(int));
So,(p + 3)is really(p + 3 * sizeof(int)).
At last,p_thirdpoints toarr[3].
Tip: You should stop coding like Java people do. Code like this (It's more readable and comprehensive) :
if() { if() { if() { } } else { } }
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