EOF 和 '\0' 的值是多少？在C中
我知道 EOF 和 '\0' 是整数类型，但如果是这样，它们不应该有一个固定值吗？
我打印了两者，EOF 为 -1，'\0' 为 0。但这些值是固定的吗？
我也有这个
int a=-1;
printf("%d",a==EOF); //printed 1
Are the value for EOF and '\0'fixed integers?
I know that EOF and '\0' are of type integers, but if so shouldn't they have a fixed value?
I printed both and got -1 for EOF and 0 for '\0'. But are these values fixed?
I also had this
int a=-1;
printf("%d",a==EOF); //printed 1
Are the value for EOF and '\0' fixed integers?
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EOF是一个宏，它扩展为类型为int的整数常量表达式和一个依赖于实现的负值，但通常为 -1。'\0'在 C++ 中是值为 0 的char，在 C 中是值为 0 的int。printf 的原因("%d",a==EOF);导致1是因为您没有将值EOF分配给a.相反，您检查a是否等于EOF并且由于这是 true (a == -1 == EOF) 它打印1..EOFis a macro which expands to an integer constant expression with typeintand an implementation dependent negative value but is very commonly -1.'\0'is acharwith value 0 in C++ and anintwith the value 0 in C.The reason why
printf("%d",a==EOF);resulted in1was because you didn't assign the valueEOFtoa. Instead you checked ifawas equal toEOFand since that was true (a == -1 == EOF) it printed1.NULL和'\0'保证计算结果为 0，因此（通过适当的转换）它们可以被认为具有相同的值；但请注意，它们代表两个截然不同的事物：NULL是 null（始终无效）指针，而'\0'是字符串终止符。EOF相反是一个负整数常量，表示流的结束；通常它是-1，但标准没有说明它的实际值。C& C++ 的不同之处在于
NULL和'\0'的类型：'\0'是char，而在 C 中它是一个int；这是因为在 C 中，所有字符文字都被视为 int。在 C++ 中
NULL“只是”一个整数 0，而在 C 中它可以定义为 0 转换为void *；这不能在 C++ 中完成（并且在注释中明确禁止），因为 C++ 在指针转换方面更加严格，void *不能隐式转换为任何其他指针类型，因此，如果NULL是一个void *，有必要在赋值时将其转换为目标指针类型：相关标准引用：
C++98/03
NULLNULL是保证计算结果为 0 的整数类型：'\0'必须存在 0 值字符：
'\0'是一个char文字：它的值为 0，因为转义序列指定了它的值：
'\0'用于终止字符串文字：EOFEOF的定义委托给 C89 标准（如第 §27.8.2“C 库文件”中所述），其中它被定义为特定于实现的负整数。C99
NULL空指针是一个 0 整数，可以选择转换为
void *；NULL是一个空指针。'\0''\0'是一个值为 0 的整数，用于终止字符串：EOFEOF是实现定义的负整数NULLand'\0'are guaranteed to evaluate to 0, so (with appropriate casts) they can be considered identical in value; notice however that they represent two very different things:NULLis a null (always invalid) pointer, while'\0'is the string terminator.EOFinstead is a negative integer constant that indicates the end of a stream; often it's -1, but the standard doesn't say anything about its actual value.C & C++ differ in the type of
NULLand'\0':'\0'is achar, while in C it's anint; this because in C all character literals are consideredints.in C++
NULLis "just" an integral 0, while in C it can be defined as a 0 casted tovoid *; this cannot be done in C++ (and it's explicitly forbidden in a note) because, being C++ more strict in pointer conversions, avoid *is not implicitly convertible to any other pointer type, so, ifNULLwas avoid *, it would be necessary to cast it to the target pointer type on assignment:Relevant standard quotations:
C++98/03
NULLNULLis an integer type guaranteed to evaluate to 0:'\0'A 0-value char must exist:
'\0'is acharliteral:and it's value is 0, since that escape sequence specifies its value:
'\0'is used to terminate strings literals:EOFThe definition of
EOFis delegated to the C89 standard (as stated in §27.8.2 "C Library files"), where it is defined as an implementation specific negative integer.C99
NULLA null pointer is a 0 integer, optionally casted to
void *;NULLis a null pointer.'\0''\0'is an integer with value 0, and is used to terminate strings:EOFEOFis an implementation-defined negative integer'\0' 始终为空字符或 0。EOF 取决于编译器，但通常为 -1，并且始终是
unsigned char无法保存的值。不要依赖 EOF 的值，因为它可以改变。始终执行 x == EOF 而不是 x == -1。 '\0' 的值始终为 0。您可以信赖这一点。'\0' is always the null character, or 0. EOF depends on the compiler, but is usually -1, and always is a value that an
unsigned charcan't hold. Don't rely on the value of EOF being anything, because it CAN CHANGE. Always do x == EOF not x == -1. The value of '\0' is ALWAYS 0. You can count on that.是的，他们是。
'\0'与NULL的值相同，均为 0（但含义不同），而 EOF 通常为 -1。在这种情况下，您会问：是 == EOF 吗？如果是 print 1（为真），则不是 print 0（为假）。
Yes, they are.
'\0'has the same value ofNULLwhich is 0 (but they mean different things), while EOF is usually -1.In this case you are asking: is a == EOF? if it is print 1 (which is true), it it's not print 0 (which is false).
'\0' 始终为 0
EOF 取决于编译器，
最常见的是 -1（在 gcc 和 g++ 中为 -1）。
'\0' is always 0
EOF is compiler dependent
most commonly its -1 (in gcc & g++ it is -1).