#include<stdio.h>#include<string.h>voidfit(char*,unsigned int);intmain(void){//字符串的优化式写法char mesg[]="Things should be as simple as possible,"" but not simpler";//修改前puts(mesg);//修改后fit(mesg,38);puts(mesg);
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{ return result / n; } return -1; } #include <stdio.h> #include <stdlib.h> #include <string.h> #include <math.h> int digPow(int n, int p) { long long sum = 0; char* s = malloc(20); sprintf(s, "%d", n); for(int i = 0; i < strlen(s); i++) { sum += pow...
itoa函数:定义:char *itoa(int value, char *string, int radix);参数:① value:需要转换的int型② string:转换后的字符串,为字符串数组③ radix:进制,范围2-36 (没run起来,一直报错,随后再补) 2. string -> int、double、long、long long atoi函数:定义:int atoi(const char *nptr);double atof(const...
代码语言:javascript 代码运行次数:0 运行 AI代码解释 #include<stdio.h>intmain(){intfun(char*s);int num;char*s="sdasdsadsad";num=fun(s);printf("%d",num);return0;}intfun(char*s){int num=0;while(*s!='\0'){num++;s++;}returnnum;}...
int main() { std::string s = "hello world"; std::cout<<s<<std::endl; for (std::string::size_type ix = 0; ix != s.size(); ++ix) s[ix] = '*'; std::cout<<"Now s is:"<<s<<std::endl; std::cout<<"s's len is:"<<s.size()<<", s[12]="<<s[100]<<std:...
1. atoi(): int atoi ( const char * str ); 说明:Parses the C stringstrinterpreting its content as an integral number, which is returned as anintvalue. 参数:str: C string beginning with the representation of an integral number.
一个int的单个字节或多或少都被你隐藏了起来。” // ENDIANNESS 端序 自测 // ENDIANNESS 端序 自测 // LICENSE: WTFPL // #include <stdio.h> #include <stdlib.h> #include <stdio.h> #include <malloc.h> #include <string.h> #include <unistd.h> static union { char c[4]; unsigned long ...
classProgram{staticvoidMain(string[] args){ Season season = Season.Autumn; Console.WriteLine($"{season}value is{(int)season}"); Console.ReadLine(); } } 使用ushort也可以作为枚举类型的基础类型。 enumSeason :ushort{ Spring, Summer, Autumn =1000, ...
#include <string> // 使用类 std::regex_iterator 来进行多次搜索. static std::string _...