scanf

What is the behavior of %(limit)[^\n] in scanf ? It is safety from overflow?

断了今生、忘了曾经 提交于 2021-02-16 05:24:44
问题 The format %(limit)[^\n] for scanf function is unsafe ? (where (limit) is the length -1 of the string) If it is unsafe, why ? And there is a safe way to implement a function that catch strings just using scanf() ? On Linux Programmer's Manual, (typing man scanf on terminal), the s format said: Matches a sequence of non-white-space characters; the next pointer must be a pointer to character array that is long enough to hold the input sequence and the terminating null byte ('\0'),which is added

understanding scanf syntax

做~自己de王妃 提交于 2021-02-11 12:47:54
问题 I have the following scanf code I am unable to understand:: char board[3][3]; int i; for(i=0;i<3;i++) scanf("%s[^\n]%*c", board[i]); Please help me understand word by word what the letters in scanf syntax mean. Thankyou. 回答1: Read a sequence of non-whitespace characters, then "[^" , newline, "]" , then one more character which is not stored anywhere. I don't think this is what actually needed. You can read scanf manpage (google it) for correct syntax. Explanation: %s - capture a sequence of

How to restrict the user to only input whole number and refuse any other characters in scanf?

最后都变了- 提交于 2021-02-08 10:20:13
问题 I'm trying to get user input and allowing them to only input a whole number which is also positive. I'm trying to figure out this task printf("Enter the first and second number in whole number: "); scanf("%d,%d", &first, &second); if (input != '\n') //notice this line I'm trying to check if is a character// printf("Error! CHARACTER NOT ACCEPTED!!\n"); else if (First <= 0) printf("Error! First must be a positive value!\n"); else if (Second <= 0) printf("Error Second must be a positive value!\n

Validating input using scanf's return value isn't working

醉酒当歌 提交于 2021-02-08 05:13:26
问题 I'm working on a tic-tac-toe final project. I ask the user to print the box number he would like to fill in, and then i use integer err to obtain the return value of scanf . Scanf should return the number of integers it has read in this case, and I am asking for one integer to be read, so as long as err != 1, it should go into the while loop. This, however, isn't working. printf("\nBox number: "); int boxNumber, err; err = scanf("%d", &boxNumber); while (err != 1) { printf("\nWrong input.");

Why can a null character be embedded in a conversion specifier for scanf?

走远了吗. 提交于 2021-02-07 12:29:18
问题 Perhaps I'm misinterpreting my results, but: #include <stdio.h> int main(void) { char buf[32] = ""; int x; x = scanf("%31[^\0]", buf); printf("x = %d, buf=%s", x, buf); } $ printf 'foo\n\0bar' | ./a.out x = 1, buf=foo Since the string literal "%31[^\0]" contains an embedded null, it seems that it should be treated the same as "%31[^" , and the compiler should complain that the [ is unmatched. Indeed, if you replace the string literal, clang gives: warning: no closing ']' for '%[' in scanf

C: scanf did not stop in infinite while loop [duplicate]

余生长醉 提交于 2021-02-07 10:58:07
问题 This question already has answers here : scanf() leaves the new line char in the buffer (4 answers) How to read / parse input in C? The FAQ (1 answer) Closed 2 years ago . I want to implement a simple code to print the error message when the type of input is not an integer. Below is a sample code. int num; while (1) { printf("Input the value of integer: "); int result = scanf(" %d", &num); if (result == 0) { printf("ERROR-Not an integer.\n"); } else if (num < 0) { printf("ERROR- Not positive.

C4473 warning for structure assignment

≯℡__Kan透↙ 提交于 2021-02-05 10:47:08
问题 I am currently working on an assignment and was curious what this warning is when compiling and how to remedy it. It will build but when I debug it will get an error screen. Below is the warning that comes up. 1>c:\users\cesteves\documents\c programming\inventory\inventory\inventory.cpp(48): warning C4473: 'scanf_s' : not enough arguments passed for format string note: placeholders and their parameters expect 2 variadic arguments, but 1 were provided note: the missing variadic argument 2 is

Why is scanf returning 0.000000 when it is supplied with a double?

久未见 提交于 2021-02-05 09:12:59
问题 I have the following assembly code (written for NASM on Linux): ; This code has been generated by the 7Basic ; compiler <http://launchpad.net/7basic> extern printf extern scanf SECTION .data printf_f: db "%f",10,0 scanf_f: db "%f",0 SECTION .bss v_0 resb 8 SECTION .text global main main: push ebp mov ebp,esp push v_0 ; load the address of the variable push scanf_f ; push the format string call scanf ; call scanf() add esp,8 push dword [v_0+4] ; load the upper-half of the double push dword [v

Why is scanf returning 0.000000 when it is supplied with a double?

不想你离开。 提交于 2021-02-05 09:09:22
问题 I have the following assembly code (written for NASM on Linux): ; This code has been generated by the 7Basic ; compiler <http://launchpad.net/7basic> extern printf extern scanf SECTION .data printf_f: db "%f",10,0 scanf_f: db "%f",0 SECTION .bss v_0 resb 8 SECTION .text global main main: push ebp mov ebp,esp push v_0 ; load the address of the variable push scanf_f ; push the format string call scanf ; call scanf() add esp,8 push dword [v_0+4] ; load the upper-half of the double push dword [v

Why is scanf returning 0.000000 when it is supplied with a double?

我的梦境 提交于 2021-02-05 09:08:52
问题 I have the following assembly code (written for NASM on Linux): ; This code has been generated by the 7Basic ; compiler <http://launchpad.net/7basic> extern printf extern scanf SECTION .data printf_f: db "%f",10,0 scanf_f: db "%f",0 SECTION .bss v_0 resb 8 SECTION .text global main main: push ebp mov ebp,esp push v_0 ; load the address of the variable push scanf_f ; push the format string call scanf ; call scanf() add esp,8 push dword [v_0+4] ; load the upper-half of the double push dword [v