gets

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

问题 This question already has answers here : Ruby: String Comparison Issues (5 answers) Closed 2 months ago . I'm a ruby beginner. I have the following code which asks the user for his name and prints it back. print 'Enter your name : ' name = gets() print("Hey,#{name} !") If I enter John Doe as the name, the output is as follows Hey,John Doe ! print unlike puts does not automatically put a new line after output but I've noticed that in the above case anything I enter after #{name} is printed on

问题 I am starting to learn programming in C language the book I am refering to code shows some source code with gets() and my IDLE recognises it as well. But still while compiling it, my compiler doesn't agree with it. Can anyone help me out? I am using gets() in the main function and using clang as the compiler. 回答1: Expanding on my comment: First, never use gets() , for any reason, even in toy code. If you see it in an example program, ignore the example and move on. It was deprecated in C99

问题 Am aware that the difference between scanf() and gets() function is that scanf() continues to read the input string until it encounters a whitespace, whereas gets() continues to read the input string until it encounters a \n or EOF(end of file). To see this difference in action, I tried writing a example on my own which is as follows: #include <stdio.h> int main() { char a[20]; printf("enter the string\n"); scanf("%s",&a); printf("the string is %s\n",a); char b[20]; printf("enter the string\n

问题 Code 1:- int main() { char str[200]; fgets(str,200,stdin); printf("%s",str); return 0; } Output:- ab cd ab cd (line feed) Code 2:- int main() { char str[200]; gets(str); printf("%s",str); return 0; } Output:- ab cd ab cd When I enter ab(space)cd(enter key) , then in case of fgets() I am getting a line feed in the output, whereas in case of gets() , no new line feed is displayed. What is the matter of the line feed in this case. 回答1: gets() and fgets() read for a FILE in to the buffer provided

问题 With the following command you can register some callback for stdin : fileevent stdin readable thatCallback This means that during the execution of the update command it will evaluate thatCallback time after time while there is input available at stdin . How can I check if input is available at stdin ? 回答1: You simply read/gets from stdin inside your callback. Basically the pattern goes like this snippet from the fileevent example from Kevin Kenny: proc isReadable { f } { # The channel is

问题 #include <iostream> using namespace std; void main(){ char name[20]; gets(name); cout<<name<<endl; } I can't found answer in google, function gets() is C or C++ language function ? Because in university I must use only C++ functions. 回答1: gets() is a C function dating to the 1960's, it does not do bounds checking and is considered dangerous, is has been kept all this years for compatibility and nothing else. Your code in valid and recommended C++ should be: #include <iostream> using namespace

问题 I just realize this 'bug' of scanf now after 8 years with C. Below scanf code will skip the leading whitespace characters from the second line of input. int x; char in[100]; scanf("%d\n",&x); gets(in); Input: 1 s x will contain 1 , but in will be just "s" not " s" Is this standard C or just gcc behaviour? 回答1: A whitespace character in your scanf format string will cause scanf to consume any (and all) white space till a non-whitespace char occurs. This seems to be standard scanf behaviour and

问题 I have this C code: #include "stdio.h" main() { struct books { char name[100],author[100]; int year,copies; }book1,book2; printf("Enter details of first book\n"); gets(book1.name); gets(book1.author); scanf("%d%d",&book1.year,&book1.copies); printf("Enter details for second book\n"); gets(book2.name); gets(book2.author); scanf("%d%d",&book2.year,&book2.copies); printf("%s\n%s\n%d\n%d\n",book1.name,book1.author,book1.year,book1.copies); printf("%s\n%s\n%d\n%d\n",book2.name,book2.author,book2

问题 Closed . This question needs details or clarity. It is not currently accepting answers. Want to improve this question? Add details and clarify the problem by editing this post. Closed 6 years ago . I wrote this: print "Enter your name:" name = gets puts "Hello #{name}. Please to meet you." and the result was like this: Hello Moemen . Pleased to meet you Why is the remainder of the string after the variable continued in another line? I want it to be "Hello Moemen. Pleased to meet you." Am I