数组和指针的困难 [英] difficulties with arrays and pointers

问题描述

我很难将十六进制值更改为整数.

在无符号char INBuffer [2]中，有一个0x0a等格式的十六进制代码.我想使用strtol函数将此值转换为long或int变量:

I''m having great difficulty changing a hex value into an integer.

In an unsigned char INBuffer[2], there is a hex code in the format 0x0a etc. I would like to convert this value into a long or int variable using the strtol function:

long int strtol ( const char * str, char ** endptr, int base );

为此，我假设我需要将INBuffer [2]中包含的值转换为字符串指针，但是无论我查找了多少示例，我似乎都无法理解指针.有人可以帮忙吗?

M.

For this I''m assuming I need to get the value contained within INBuffer[2] into a string pointer, but no matter how many examples I look up i just can''t seem to get my head around pointers. Could somebody please help?

M.

推荐答案

在我看来有点困惑:具有两个元素的无符号char缓冲区不能保存格式为"0x0a"的十六进制代码是四个字符.

如果您的意思是两个无符号字符中都有一个等于十六进制值0A的值，并且其范围介于0000和FFFF十六进制之间，那么您需要执行以下操作:

That seems a little confused to me: an unsigned char buffer with two elements can''t hold a hex code in the format "0x0a" since that is four characters.

If you mean that there is a value in the two unsigned chars that equates to the hex value 0A, and that it has a range between 0000 and FFFF hex, then you need to do this:

int i = ((int) INBuffer[0] & 0xFF) + (((int) INBuffer[1] << 8) & 0xFF00);

(那里有AND可以防止符号扩展，无论如何都不应发生).您可能需要根据数据格式交换数组索引:big-endian或little-little-endian

如果您的意思是INBuffer包含两个分别为"0"和"A"的字节，那么您需要将它们复制到一个足以容纳三个字节的字符数组中，并使其成为以null结尾的字符串:

(The ANDs are there to prevent sign extentions, which shouldn''t happen anyway) You may need to swap round the array indexes depending on your data format: big- or little-endian

If you mean that INBuffer contains two bytes which are ''0'' and ''A'' respectively, then you need to copy them to a character array big enough to hold three bytes, and make it a null terminated string:

unsigned char data[3];
data[0] = INBuffer[0];
data[1] = INBuffer[1];
data[2] = '\0';

然后可以使用它将其转换为int:

You can then use that to convert it to a int:

char * pEnd;
long int i = strtol(data, &pEnd, 16);

我对您的问题感到困惑，但让我们尝试一下...

您有一个未签名的char INBuffer [2].这意味着它是一个两个字符的缓冲区.可以使用INBuffer [0]和INBuffer [1]分别访问此缓冲区的两个元素.

如果要在INBuffer [0]中包含十进制值，则可以执行此操作

I''m confused by your question, but let''s try this...

You have an unsigned char INBuffer[2]. This means it''s a two character buffer. The two elements of this buffer can be accessed individually by using INBuffer[0] and INBuffer[1].

If you want the decimal value contained in INBuffer[0], you can do this

int buf0val = (int)INBuffer[0];

同样，如果您希望该缓冲区的第二个字符包含该值，则可以执行此操作

Likewise if you want the value contained in the second character of this buffer, you can do this

int buf1val = (int)INBuffer[1]

但是，如果您的意思是第一个字符包含ASCII数字0，第二个字符包含ASCII"A"，则您遇到了另一个问题.

您不能对此使用strtol，因为此(两个字符)字符串不是NULL终止的-就像上面的OriginalGriff所说的那样.

希望对您有所帮助.

However if you mean that the first character contains the ASCII number 0 and the second character contains the ASCII ''A'', the you have a different problem.

You can''t use strtol on this, because this (two character) string is not NULL terminated - just as OriginalGriff said above.

Hope that helps.

我认为您可以这样做，

I think You can do like this,

char*chVal  = "0x0a";
cout<<"size:: "<<sizeof(chVal)<<endl;
long lVal = strtol(chVal, NULL, 16);

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