unsigned

问题 In C program, if an integer is signed integer , then the highest bit is 1 , otherwise 0 . Lets take char and unsigned char , the range for a signed char is -128 to 127 and unsigned char is 0 to 255 , but in fact their hexadecimal are in range of 0x00 to 0xff . My question is now if a char and unsigned char are stored in memory using 8 bits binary number, how does the computer itself know whether it is signed or unsigned ? char a = 0xff; printf("%d", a); //its result is -1. unsigned char a =

问题 This question already has answers here : Declaring an unsigned int in Java (11 answers) Closed last year . I want to assign 4294967295 to a variable ( 2^32-1 ) It is obvious that I can't do that with Integer , and can do it with Long . However, I noted that Java 8 offers Unsigned Integers (at least some methods). Does any one know what the method, Integer.parseUnsignedInt() does? When I input "4294967295" to that, and print the variable, it gives the output as -1 ( -2 for 4294967294 , -3 for

问题 I get very confused when it seems like sometimes my code treats a certain value as signed and sometimes it treats it as unsigned when comparing values. How does the code know whether a value is signed or unsigned? 回答1: Why do you think that assembly code has to "know" if a value is signed or unsigned? For most operations the results of a signed and an usigned operation are the same: signed int a = 5; signed int b = -6; // 0xFFFFFFFA signed int c; c = a + b; // results in -1 which is

问题 unsigned int i; for (i = 100; i <= 0; --i) printf("%d\n",i); 回答1: Should be i >= 0 in the second condition in the loop if you want it to loop from 100 to 0. That, and as others have pointed out, you'll need to change your definition of i to a signed integer (just int ) because when the counter is meant to be -1, it will be some other positive number because you declared it an unsigned int . 回答2: Since i is unsigned, it will never be less than zero. Drop unsigned . Also, swap the <= for >= .

问题 Consider the following piece of C code: #include <stdint.h> uint32_t inc(uint16_t x) { return x+1; } When compiled with gcc-4.4.3 with flags -std=c99 -march=core2 -msse4.1 -O2 -pipe -Wall on a pure x86_64 system, it produces movzwl %di,%eax inc %eax retq Now, unsigned overflow is predicted in C. I do not know much about x86_64 assembly, but as far as I can see the 16bit argument register is being moved to a 32bit register, which is incremented and returned. My question is, what if x == UINT16

问题 I am having problem with an output error in a waveform, basically my code works as a counter, when i have a load signal equal '1' the counter goes up, if the load signal is '0' the counter doesn't counts. I have a clear signal to get the counter in 0, my problem is in the output, the output shows always the same value and doesn't get in 0 when the clear signal is equal 1. Below the waveform: Below the code: library ieee; use ieee.std_logic_1164.all; use ieee.numeric_std.all; entity tot is

问题 I have a pointer to a buffer of bytes from which I am copying every even indexed bytes to an int(because of the protocol that the data is stored into buffer I know the odd cycles are for read). Now when I do this signed int a; ... //inside a loop a = buffer[2*i]; //buffer is unsigned It gives me an unsigned number. However when I do this a = (int8_t)buffer[2*i] the number is presented in signed form. That is forcing me to rethink how sign extension in c work, especially in scenarios like

问题 I have an INT(3) UNSIGNED column. If I insert a value with character length more that 3, it doesn't clip that value but inserts it. Whats happening? 回答1: FROM What does "size" in int(size) of MySQL mean? Finally, let's come to the place of the manual where there is the biggest hint to what the number means: Several of the data type descriptions use these conventions: M indicates the maximum display width for integer types. For floating-point and fixed-point types, M is the total number of

问题 I've been trying to convert a hexadecimal number saved in a buffer to an unsigned int. However the "0x00" in front of every hexadecimal number that I'm reading from has been giving me problem, in essence the problem (in a downscaled version) looks like this: char b[] = "0x0014A12"; std::stringstream ss; unsigned int i; ss << std::hex << b; ss >> i; cout << i << endl; Any tips? Note: The program outputs a high decimal nubmer which equals CCCCCC in hex. 回答1: This works fine for me: #include

问题 I have to implement two functions that convert signed to unsigned and vice versa. I am using c++11 and Linux. The system is two's complement and can take char, int, long etc.. The interface must be as stated and I have tried to implement something. Are there better ways to do this? Are these correct? How can I change the implementation based on the number of bits? I need some advice. uint32_t signedToUnsigned(int32_t x, uint8_t bits) { return ( x > 0 ? x:-x); } int32_t unsignedToSigned(uint32