如何在C中为2D数组位置赋值?

So I'm quite a newbie, so please go soft.

我是个新手，所以请温柔点。

I have created a 2D array, 60x30 and want to display it on the screen as a grid, by doing a double for loop. I am using a simple character '.' for each slot of the grid just for a test.

我创建了一个2D数组，60x30，并希望通过双for循环在屏幕上以网格的形式显示它。我用的是一个简单的字符。“这是为了测试网格的每个槽。”

char FrameBuffer[29][59];
for (int i = 0; i <= 29; i++) 
{
    for (int j = 0; j <= 59; j++)
    {
        FrameBuffer[i, j] = '.';
        printf("%c ", FrameBuffer[i,j]);
    }
    printf("\n");
}

However, whenever I try and assign a value to a position in my 2D array, e.g.

但是，每当我尝试为我的2D数组中的一个位置赋值时，例如。

FrameBuffer[0,1] = '.',

I am greeted with an error:

我犯了一个错误:

Expression must be a modifiable lvalue

3 个解决方案

#1

FrameBuffer[29][59] This is a 29x59 array, you need a 30x60.

这是一个29x59数组，需要30x60。

Please don't mix up array declarations with array indexing. Array declarations are fully sane; if you need 30x60 then you type FrameBuffer[30][60]. When you access the array however, you start at index 0.

请不要将数组声明与数组索引混淆。数组声明完全正常;如果需要30x60，则输入FrameBuffer[30][60]。然而，当您访问数组时，您从索引0开始。

Simply change the code to:

只需将代码更改为:

char FrameBuffer[30][60];
for (int i = 0; i < 30; i++)
{
  for (int j = 0; j < 60; j++)
  {
    FrameBuffer[i][j] = '.'; // note the correct syntax here

EDIT

编辑

As a curious side-effect, FrameBuffer[i, j] is interpreted as something entirely different than intended. The comma here is regarded as the comma operator, a special kind of operator which evaluates the left expression i, then the right expression j, then returns j.

作为一个奇怪的副作用，FrameBuffer[i, j]被解释为完全不同于预期的东西。这里的逗号被视为逗号运算符，是一种特殊的运算符，它计算左表达式i，然后计算右表达式j，然后返回j。

Meaning that the code ended up completely equivalent to FrameBuffer[j] = '.'. Where FrameBuffer[j] is a whole array, not a char. You can't assign a value to an array this way, an array is not a "lvalue", which explains the compiler error text.

这意味着代码最终完全等同于FrameBuffer[j] = '.'。其中FrameBuffer[j]是一个完整的数组，而不是字符。不能这样给数组赋值，数组不是“lvalue”，这解释了编译器错误文本。

#2

FrameBuffer[i,j] = '.';

FrameBuffer(i,j)= '。';

printf("%c ", FrameBuffer[i,j]);

printf(" % c”,FrameBuffer(i,j));

modify these two statements to

将这两个语句修改为

FrameBuffer[i][j] = '.';

printf("%c ", FrameBuffer[i][j]);

[i,j] means nothing to the compiler. It's nothing but an invalid syntax.

[i,j]对编译器没有任何意义。它只是一个无效的语法。

also correct you array.

还正确的数组。

FrameBuffer[29][59]it is an array of row=29 and col=59

FrameBuffer[29][59]它是一行=29和col=59的数组。

change it to FrameBuffer[60][30];

将其更改为FrameBuffer[60][30];

#3

-1

char FrameBuffer[31][61];
for (int i = 0; i <= 29; i++)
            {
                for (int j = 0; j <= 59; j++)
                {
                    FrameBuffer[i][j] = '.';
                    printf("%c ", FrameBuffer[i][j]);
                }
                FrameBuffefr[i][j] = '\0';
                printf("\n");
            }
   FrameBuffer[i] = NULL;
   }

Here you assign 30*60 characters, so you need to allocate 30 * 60 + 1 on each to add a character to indicate the end ('\0'). (easy to parse it, because you don't need the size you can do a while(Framebuffer[i][j]) it will be a value 0 on the '\0').

在这里，您分配了30*60个字符，因此您需要为每个分配30*60 + 1以添加一个字符来指示结束('\0')。(解析起来很容易，因为您不需要一次性使用的大小(Framebuffer[i][j]))它在'\0'上的值为0)。

#1