signed

问题 I would like to know how I could get the original content from a signed pdf document using iText java library or another one. Thanks UPDATE 1: Possible example: PdfReader reader = new PdfReader(PATH_TO_PDF); AcroFields fields = reader.getAcroFields(); ArrayList<String> signatures = fields.getSignatureNames(); for (String signature : signatures) { // Start revision extraction ByteArrayOutputStream out = new ByteArrayOutputStream(); byte bb[] = new byte[8192]; InputStream ip = fields

问题 I'm reading C++ Primer and I'm slightly confused by a few comments which talk about how Bitwise operators deal with signed types. I'll quote: Quote #1 (When talking about Bitwise operators) "If the operand is signed and its value is negative, then the way that the “sign bit” is handled in a number of the bitwise operations is machine dependent. Moreover, doing a left shift that changes the value of the sign bit is undefined" Quote #2 (When talking about the rightshift operator) "If that

问题 This question already has answers here : Signed vs. unsigned integers for lengths/counts (4 answers) Closed 6 years ago . At my code, I do not use int or unsigned int. I only use size_t or ssize_t for portable. For example: typedef size_t intc; // (instead of unsigned int) typedef ssize_t uintc; // (instead of int) Because strlen , string , vector ... all use size_t , so I usually use size_t . And I only use ssize_t when it may be negative. But I find that: The unsigned integer types are

问题 This question might be very basic but i post here only after days of googling and for my proper basic understanding of signed integers in C. Actually some say signed int has range -32767 to 32767 and others say it has range -32768 to 32767 Let us have int a=5 (signed / let us consider just 1 byte) *the 1st representation of a=5 is represented as 00000101 as a positive number and a=-5 is represented as 10000101 (so range -32767 to 32767 justified) (here the msb/sign bit is 1/0 the number will

问题 I need to implement a 32-bit arithmetic right shift from logical shifts, and, or, xor and normal integer arithmetic operations. I read somewhere the following is supposed to work: (x>>N)|(((1<<N)-1)<<(32-N)) x is the integer that will be shifted and N is the amount of bits to shift. This works for negative (msb is 1) numbers but not for positive numbers (msb is 0). Does anyone know an efficient algorithm that always produces the right result? 回答1: You can use this (x >> N) | (-(x < 0) << (32

问题 Is there any C compiler which takes the default type of char as unsigned unless explicitly mentioned by the user in the file or project settings? /Kanu_ 回答1: GCC does. But only when compiling for platforms where an unsigned char is the convention, including ARM linux[*]. When GCC compiles for x86, the default is for char to be signed. [*] Or at least it has been in the past. For all I know linux has switched to a different default ABI on ARM since. Update '2013: ARM compilers (gcc, clang) for

问题 I know that addition of two unsigned integers larger than the bus size of a given processor can be achieved via the carry flag. And normally, the same is true for signed integers using the overflow flag. However, the Intel 8085 only possesses a Sign flag, not an Overflow flag, so how does it deal with signed integer arithmetic? 回答1: As you know, the overflow flag is only relevant for signed integer arithmetic. On processors whose ALU has both overflow and carry flags (like x86), both of these

问题 I want to compare two 16-bit numbers and branch on the result: the equivalent of if (a<b) goto negative . I'm using an Intel 8080. The Z80 has a signed arithmetic overflow flag which can be used for this with some degree of effort. The standard code is: ld de, _left ld hl, _right ld a, e sub a, l ld a, d sbc a, h jp po, $+5 ; branch on overflow flag not set xor a, 0x80 ; flip sign bit jm negative ; actually do the test But the 8080 isn't a strict subset of the Z80, and the code above won't

问题 Given that signed and unsigned ints use the same registers, etc., and just interpret bit patterns differently, and C chars are basically just 8-bit ints, what's the difference between signed and unsigned chars in C? I understand that the signedness of char is implementation defined, and I simply can't understand how it could ever make a difference, at least when char is used to hold strings instead of to do math. 回答1: It won't make a difference for strings. But in C you can use a char to do

问题 I have just started learning C and a question has bugged me for a while now. If I write int i = -1; unsigned int j = 2; unsigned int k = -2; What is the type of integer literal -1 and 2 and -2 , and how does it get converted to get stored in signed int and unsigned int ? What is meant by signed integer, is that the property of variable or integer literal too? Like -2 is signed integer and 2 is unsigned integer? 回答1: First off, -1 is not an integer constant. It's an expression consisting of a