stdio

问题 So I'm working on a program which needs to format output. The output is supposed to be aligned, and it does do so with small numbers: But then when I give big numbers, it no longer works: My code is really but here's the part that prints the main output: /* The following code prints out the data */ printf("\n\nStatistics: \n\n"); printf("Descrip\t\tNumber:\t\tTotal:\t\tAverage:\n\n"); printf("Normal\t\t%d\t\t%d\t\t%d\n\n",normal_counter,normal_total,normal_average); printf("Short\t\t%d\t\t%d

问题 I don't understand what the buffer is doing and how it's used. (Also, if you can explain what a buffer normally does) In particular, why do I need fflush in this example? int main(int argc, char **argv) { int pid, status; int newfd; /* new file descriptor */ if (argc != 2) { fprintf(stderr, "usage: %s output_file\n", argv[0]); exit(1); } if ((newfd = open(argv[1], O_CREAT|O_TRUNC|O_WRONLY, 0644)) < 0) { perror(argv[1]); /* open failed */ exit(1); } printf("This goes to the standard output.\n"

问题 Problem Docker image sizes should commonly be as small as possible. Using full-blown environments like a standard python image results often, with all dependencies installed, in heavily bloated images. Packaging python into stand-alone executables (e.g. using pyinstaller) is a perfect way of reducing image sizes and overall complexity. Environment : python3.6 , pyinstaller==3.4 The problem arising is, that python uses per default buffered stdio. This can be mitigated by running python scripts

问题 Problem Docker image sizes should commonly be as small as possible. Using full-blown environments like a standard python image results often, with all dependencies installed, in heavily bloated images. Packaging python into stand-alone executables (e.g. using pyinstaller) is a perfect way of reducing image sizes and overall complexity. Environment : python3.6 , pyinstaller==3.4 The problem arising is, that python uses per default buffered stdio. This can be mitigated by running python scripts

问题 Problem Docker image sizes should commonly be as small as possible. Using full-blown environments like a standard python image results often, with all dependencies installed, in heavily bloated images. Packaging python into stand-alone executables (e.g. using pyinstaller) is a perfect way of reducing image sizes and overall complexity. Environment : python3.6 , pyinstaller==3.4 The problem arising is, that python uses per default buffered stdio. This can be mitigated by running python scripts

问题 I'm learning Rust, and I'm somewhat stumped. I'm trying to give the user the option of writing output to stdout or to a supplied filename. I started with the example code that's given for using extra::getopts located here. From there, in the do_work function, I'm trying to do this: use std::io::stdio::stdout; use std::io::buffered::BufferedWriter; fn do_work( input: &str, out: Option<~str> ) { println!( "Input: {}", input ); println!( "Output: {}", match out { Some(x) => x, None => ~"Using

问题 I am aware that most terminals are by default in line buffer mode. i.e. output is buffered and not directed to stdout until a new-line character is encountered. So I would expect this to print nothing (at least before the buffer is filled up): int main() { while(1) { printf("Haha"); sleep(1); } return 0; } It indeed prints nothing for a short period of time. If I want to print "Haha" every second, I can either printf("Haha\n") or do fflush(stdout) after printf. (I know this is not so portable

问题 I am aware that most terminals are by default in line buffer mode. i.e. output is buffered and not directed to stdout until a new-line character is encountered. So I would expect this to print nothing (at least before the buffer is filled up): int main() { while(1) { printf("Haha"); sleep(1); } return 0; } It indeed prints nothing for a short period of time. If I want to print "Haha" every second, I can either printf("Haha\n") or do fflush(stdout) after printf. (I know this is not so portable

问题 This question already has answers here : What is the effect of trailing white space in a scanf() format string? (4 answers) Closed 3 years ago . I just had a test in my C class today and I have reason to believe that answer might be incorrect. scanf("%d\n", &x); // Evaluate the expression for the string "54321\n" The idea is pretty simplistic. Find an integer and place the scanned number at the memory location corresponding with integer variable x . However, I don't believe that this call to

问题 I'm confused as to how the writing processing goes in C. So I have a string, s, that I want to write to the output. To do that, I use fputs: fputs(s, stdout); But apparently this does not write to the output, but merely collect the data for writing? Where exactly is it collected? So I have to wait until the program exits or till I call fflush() till the output is actually written into stdout? Am I right? 回答1: The C Standard IO streams are operating in one of three modes: fully buffered line