C语言 如何使用十六进制表示法为 char* 赋值？
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How to assign a value to a char* using hex notation?
提问by Andrew-Dufresne
I usually use pointers in the following manner
我通常以下列方式使用指针
char *ptr = malloc( sizeof(char) * 100 );
memset( ptr, 0, 100 ) ;
strncpy( ptr, "cat" , 100 - 1 );
But this time instead of using "cat", I want to use it ASCII equivalent in hex.
但这一次，我想使用与 ASCII 等效的十六进制形式，而不是使用“cat”。
cat = 0x63, 0x61, 0x74, 0x00
猫 = 0x63、0x61、0x74、0x00
I tried
我试过
strncpy( ptr, "0x630x61" , 100 - 1 );
But it fails as expected.
但它按预期失败。
What is the correct syntax?
什么是正确的语法？
Do I need to put a 0x00 too? For a moment lets forget about memset, now do I need to put a 0x00? Because in "cat" notation, a null is automatically placed.
我也需要输入 0x00 吗？让我们暂时忘记memset，现在我需要输入 0x00 吗？因为在“cat”符号中，会自动放置一个空值。
Regards
问候
回答by Claudiu
\xXXis the syntax for inserting characters in hex format. so yours would be:
\xXX是以十六进制格式插入字符的语法。所以你的将是：
strncpy( ptr, "\x63\x61\x74", 100 - 1);
You don't need to put in a \x00since having quotes automatically null-delimits the string.
您不需要输入 a\x00因为引号会自动对字符串进行空分隔。
回答by Martin Beckett
Note, you only need \ inside the " " string
请注意，您只需要在 " " 字符串中使用 \
char cat[4];
cat[0] = 0x63;
cat[1] = 0x61;
cat[2] = 0x74;
car[3] = 0x00;
char cat[] = "\x63\x61\x74"; // note the strncpy( ptr, "\x63\x61" , 100 - 1 );
is added for you
char cat[] = { 0x63, 0x61, 0x74, 0x00 };
Are all the same
都一样
回答by Federico A. Ramponi
0x63is an integer hexadecimal literal; The C compiler parses it as such within code. But within a string it is interpreted as a sequence of characters 0,x,6,3.
The literal for the charwith value 63 hex. is '\x63', and within strings you must use this notation.
"c\x63"is the literal for a zero-terminatedstring, irrespective of the characters within the quotes (or of the notation by which you denote them), so no, you don't need to append a trailing zero manually.
0x63是一个整数十六进制文字；C 编译器在代码中解析它。但在一个字符串中，它被解释为一个字符序列0,x,6,3。char值为 63 十六进制的文字。is '\x63'，并且在字符串中您必须使用此表示法。
"c\x63"是零终止字符串的文字，与引号内的字符（或表示它们的符号）无关，因此不，您不需要手动附加尾随零。
