将整数值存储为C中的字符串。-StoreintegervaluesasastringinC

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When you write

当你写

char* hexVal = "";

you are setting hexVal to point to a string literal, later in your code you try to strcat to that address which will cause undefined behavior.

您正在设置hexVal以指向一个字符串字面量，稍后在您的代码中，您尝试将其strcat转换到将导致未定义行为的地址。

What you need to do is to allocate a large enough area to hold your resulting string and then let hexVal point to that.

你需要做的是分配一个足够大的区域来容纳你的结果字符串，然后让hexVal指向那个。

E.g.

如。

char* hexVal = malloc(100); // or how much bytes you need

then just do

然后就做

strcat(hexVal, temp);

alt. allocate on stack

alt.分配在堆栈

char hexVal[100];

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You are approaching this wrong. In general, if you have an int i then you can't just typecast a char *cp to the address (or the value) of i and expect it to magically become a string that you can printf or use in strcat. For one thing, strings are null-terminated and don't have a fixed length, while ints have a fixed size of typically 32 bits long.

你正接近这个错误。一般来说，如果您有一个int i，那么您不能仅仅将char *cp类型转换为i的地址(或值)，并期望它会神奇地成为您可以在strcat中打印或使用的字符串。首先，字符串以null结尾，并且没有固定长度，而ints的长度通常为32位。

You have to create a separate buffer (memory space) where snprintf will happily create a string-representation of your int value for you.

您必须创建一个单独的缓冲区(内存空间)，snprintf将愉快地为您创建int值的字符串表示。

I think that your question is more about understanding how programming, pointers and C work in general, than about ints and strings and their conversion.

我认为你的问题更多的是关于如何理解编程，指针和C的一般工作，而不是关于int和字符串以及它们的转换。

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You are getting undefined behavior since there is no writable memory at hexVal, which just points at a read-only area containing a character with the value 0. It is a valid string, but it's constant and has length 0, you cannot append to it.

您正在获得未定义的行为，因为在hexVal中没有可写的内存，它只指向包含具有值0的字符的只读区域。它是一个有效的字符串，但是它是常量，长度为0，你不能附加到它。

Also, of course you can't cast an integer into a "string", that's just not how C works. This:

当然，你也不能把整数转换成字符串，这不是C的工作方式。这样的:

temp=(char*) val;

simply re-interprets the value of val as a pointer (i.e. as an address), it doesn't magically compute the proper sequence of digit characters used to represent the address, i.e. it doesn't convert val to a string.

简单地将val的值重新解释为一个指针(即一个地址)，它不会神奇地计算出用于表示地址的数字字符的正确序列，也就是说它不会将val转换为字符串。

You can use snprintf() to convert an integer to a string.

可以使用snprintf()将整数转换为字符串。

So, to summarize:

所以,总结一下:

• Change hexval's declaration into e.g. char hexval[32] = "";, this declares it as an array of 32 characters, giving you plenty of space into which to build a number as a string. It also initializes the array so the first character is 0, thus making it into an empty string (with space to grow).
• 将hexval的声明改为:char hexval[32] = ";，这声明它是一个32个字符的数组，给了您足够的空间来构建一个数字作为字符串。它还初始化数组，因此第一个字符是0，从而使它变成一个空字符串(有空间增长)。
• Use e.g. snprintf(tmp, sizeof tmp, "%d", 4711); to format a number into a string (which should be e.g. char tmp[16];) in decimal form.
• 使用snprintf(tmp, sizeof tmp，“%d”，4711);将一个数字格式化成一个字符串(应该是char tmp[16];)以十进制形式。
• Use strcat(hexval, tmp); to concatenate the newly built numeric string onto hexval.
• 使用strcat(hexval、tmp);将新建的数字字符串连接到hexval。
• BEWARE that you can't concatenate forever, you will run out of space if you do it too long. Adjust the sizes, in that case.
• 要注意的是，如果你做的时间太长，你的空间就会耗尽。调整大小，在这种情况下。
• Check return values where possible, read the manual pages (Google man FUNCTION where FUNCTION is a standard C library function like strcat).
• 在可能的情况下检查返回值，阅读手册页(函数是标准的C库函数，如strcat)。