atoi | 易学教程

C strings behavior, and atoi function

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问题 I wonder why the two values of int don't validate the if condition even if it is true. printf shows both of them are equal. Is buffer overflow able to affect the behavior of if conditions,corrupting other code sections behavior. #include <stdio.h> #include <stdlib.h> #include <time.h> int main(void) { srand(time(NULL)); char instring[2]; // when this increases somehow I get the right behavior int inint; int guess; guess = rand() % 127; inint = ~guess; printf("%i\n", guess); //testing with

C character values arithmetic

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问题 I have been reading from the book "The C Programming Language" learning C, and I stumbled upon the arithmetic s[i] - '0' which they said that it gives the numeric value of the character stored in s[i]. I didn't quite understand it, how could it give the value by subtraction? Note This is used in the atoi function, which converts a string of digits into its numeric equivalent. Thanks 回答1: Basically, what you need to understand is that on a modern computer, all information is stored digitally

Is there an atoi implementation for C#

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问题 I have a file with numbers saved as strings. I am looking to load this data as a char[] then convert this array to an integer value. I have several values to load and am trying to avoid the overhead of creating thousands of temp strings.. 回答1: Yup int.Parse(string) If you absolutely must convert the strings to character arrays (not sure why), you can use: int.Parse(new string(myCharArray)); On a further note... I am looking to load this data as a char[] then convert this array to an integer

Why does string to int conversion not print the number 0 if its at first index [duplicate]

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问题 This question already has answers here : Keep Leading zeros C (2 answers) keeping leading zeros in C [duplicate] (3 answers) Closed last year . I am wondering as to why when converting strings to int with either atoi or strtol it does not print the number 0 if its the first index, take this code for example char s[] = "0929784"; long temp = strtol(s, NULL, 10); printf("%li", temp); OUTPUT: 929784 is there a way to have the 0 print? 回答1: Use char s[] = "0929784"; size_t len = strlen(s); long

What is difference between my atoi() calls?

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问题 I have a big number stored in a string and try to extract a single digit. But what are the differences between those calls? #include <iostream> #include <string> int main(){ std::string bigNumber = "93485720394857230"; char tmp = bigNumber.at(5); int digit = atoi(&tmp); int digit2 = atoi(&bigNumber.at(5)) int digit3 = atoi(&bigNumber.at(12)); std::cout << "digit: " << digit << std::endl; std::cout << "digit2: " << digit2 << std::endl; std::cout << "digit3: " << digit3 << std::endl; } This

C Convert String to Ints Issue

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问题 I'm trying to parse some input on an embedded system. I'm expecting something like this: SET VARNAME=1,2,3,4,5,6,7,8,9,10\0 When I'm converting the separate strings to ints, both atoi() and strtol() seem to be returning 0 if the string begins with 8. Here is my code: char *pch, *name, *vars; signed long value[256]; int i; #ifdef UARTDEBUG char convert[100]; #endif if(strncmp(inBuffer, "SET",3)==0) { pch = strtok(inBuffer," "); pch = strtok(NULL," "); name = strtok(pch, "="); vars = strtok

K&R atoi-general memory leak

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问题 I am following K&R second edition examples to learn C and coding as I think this is correct way of doing things. Anyhow when I run this program post compilation the program get stuck. I used valgrind for execution of compiled script. #include <ctype.h> int atoi(char s[]) { int i, n, sign; for(i = 0; isspace(s[i]); i++) ; sign = (s[i] == '-')? -1 : 1; if (s[i] == '+'|| s[i] == '-') i++; for (int n = 0; isdigit(s[i]); n++) n = 10 * n + (s[i]-'0'); return sign * n; } int main() { char input[] =

c atoi() for wide chars on linux?

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问题 Is there a c atoi() equivalent for wide chars on Linux? I can find something for MS (wtoi) but I can find anything in a standard Linux lib. 回答1: You can use wcstol to convert from wide strings to integer values. 回答2: It is unusual for a Linux program to use wchar_t type. The reason being that Linux uses utf-8 as the standard encoding. char const* strings are assumed to be utf-8 strings by glibc. Ascii digits and utf-8 digits have the same byte representation, so that atoi() works both on

题8、字符串转换整数 (atoi)

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题8、字符串转换整数题目思路代码题目请你来实现一个 atoi 函数，使其能将字符串转换成整数。首先，该函数会根据需要丢弃无用的开头空格字符，直到寻找到第一个非空格的字符为止。当我们寻找到的第一个非空字符为正或者负号时，则将该符号与之后面尽可能多的连续数字组合起来，作为该整数的正负号；假如第一个非空字符是数字，则直接将其与之后连续的数字字符组合起来，形成整数。该字符串除了有效的整数部分之后也可能会存在多余的字符，这些字符可以被忽略，它们对于函数不应该造成影响。注意：假如该字符串中的第一个非空格字符不是一个有效整数字符、字符串为空或字符串仅包含空白字符时，则你的函数不需要进行转换。在任何情况下，若函数不能进行有效的转换时，请返回 0。说明：假设我们的环境只能存储 32 位大小的有符号整数，那么其数值范围为 [−231, 231 − 1]。如果数值超过这个范围，qing返回 INT_MAX (231 − 1) 或 INT_MIN (−231) 。示例 1: 输入: “42” 输出: 42 示例 2: 输入: " -42" 输出: -42 解释: 第一个非空白字符为 ‘-’, 它是一个负号。我们尽可能将负号与后面所有连续出现的数字组合起来，最后得到 -42 。示例 3: 输入: “4193 with words” 输出: 4193 解释: 转换截止于数字

atoi,itoa,strcpy, strcmp,strcpy, strcpy_s, memc...

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strcpy()、strlen()、memcpy()、memmove()、memset()的实现 strcpy(), 字符串拷贝. char * strcpy( char * strDest, const char * strSrc) { assert((strDest != NULL) && (strSrc != NULL)); char * address = strDest; while ( ( * strDest ++ = * strSrc ++ ) != ' \0 ' ) NULL ; return address ; } strlen, 第一种方法： int strlen( const char *str) { assert(str != NULL); int len = 0; while ((*str++) != '\0' ) len++; return len; } 第二种方法： int strlen( const char *str) { assert(str != NULL); const char *p = str; while ((*p++) != '\0' ); return p - str - 1; } 第三种方法： int strlen( const char * str) { if (str[0] == '\0' ) return 0; else