signed

问题 C program to find nth power of integer m without pow(). Input: m=3 n=2 output: 9.000 Tests to validate the program works as expected! For negative M Input : -2 3 output : -8.000 For negative N Input : 2 -3 output : 0.125000 For negative M and N Input : -2 -3 output : -0.125000 However I am not getting the desired output void main() { signed int m, n; int i; float p; clrscr(); printf("Enter the number and its power (exponent)\n"); scanf("%d%d",&m,&n); p=1; if (n==0) { printf("%d raised to %d

问题 I have following function which counts the number of binary digits in an unsigned 32-bit integer. uint32_t L(uint32_t in) { uint32_t rc = 0; while (in) { rc++; in >>= 1; } return(rc); } Could anyone tell me please in case of signed 32-bit integer, which approach i should take ? implementing two's complement is an option. if you have any better approach, please let me know. 回答1: What about: uint32_t count_bits(int32_t in) { uint32_t unsigned_in = (uint32_t) in; uint32_t rc = 0; while (unsigned

问题 I would like to pass a signed int to gsl_rng_uniform_int (const gsl_rng * r, unsigned long int n) . The signed int that I'm passing is greater than or equal to zero. The function will return a number between 0 and n , so if I pass it a positive signed int, it will return something that is in the range of a signed int. I then want to store the return value in a signed int. What's the cleanest way to do this with the obvious expected behaviour? I'm on a 64-bit compiler on a 64-bit Linux machine

问题 I have this method that converts a signed or non-signed byte to int , but it doesn't return what it is supposed to return. Can someone point out the issue in the below code? public int convertByteToInt(byte[] b){ int value= 0; for(int i=0;i<b.length;i++){ int n=(b[i]<0?(int)b[i]+256:(int)b[i])<<(8*i); value+=n; } return value; } Edited : I'am actualy reading a wav file in order to calculate the SNR. the returned value from the conversion should give something beetween 0 and 255. The

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 I'm making a class PackedUnsigned1616 which stores two unsigned shorts in one int, and a class PackedSigned1616 which stores two signed shorts in one int. I've read up on bitwise operations, but I'm still confused on how to deal with signed and unsigned and values that are larger or smaller that a short's range (they are passed in as two ints). Here's what I've got so far: public final class PackedUnsigned1616 { public final int field; private static final int RIGHT = (2 << 15) - 1; private

问题 I want to convert a string into a signed int. Following is the requirement. I have stored hex value as a string in buffer. Now I want to convert that value into signed int. buf = "fb869e" Convert this into signed int. So o/p should be -293218. but when I'm trying to convert using strtol I'm getting 16483998. So what I should I do? 回答1: The hexadecimal number 0xfb869e is not negative. The inbuilt number conversion functions will not convert it to a negative value, since its value is positive.

问题 I have an issue in the mind and that is since the jump instruction changes EIP register by adding signed offsets to it(if I'm not making a mistake here), on IA-32 architecture how would going upward in memory from location 0x7FFFFFFF(biggest positive number in signed logic) to 0x80000000(least negative number in signed logic) be possible? or maybe there shouldn't be such jump due to the nature of signed logic? 回答1: Signed and unsigned are just two ways of interpreting the same bit pattern.

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 want to generate a signed apk. But I reached the 64k methods limit so I searched for some solutions and one of them is to use proguard. I edited my build.gradle like this: buildTypes { release { minifyEnabled true shrinkResources true proguardFiles getDefaultProguardFile('proguard-android.txt'), 'proguard-rules.pro' } } When I then try to generate the signed apk I get a lot of warnings: Warning:com.akexorcist.roundcornerprogressbar.TextRoundCornerProgressBar$2: can't find referenced method