fwrite

write XImage to .bmp file in C

旧街凉风 提交于 2020-01-03 20:11:43
问题 I am writing an XImage to a file "bitmap0.bmp" using the following code but when i open the newly created file with imageViewer in fedora it gives "Premature end of file" error and does not display anything. could anyone please point out the problem in the following code? typedef struct tagBITMAPFILEHEADER { WORD bfType; DWORD bfSize; WORD bfReserved1; WORD bfReserved2; DWORD bfOffBits; } BITMAPFILEHEADER; typedef struct tagBITMAPINFOHEADER{ DWORD biSize; LONG biWidth; LONG biHeight; WORD

write XImage to .bmp file in C

大城市里の小女人 提交于 2020-01-03 20:11:08
问题 I am writing an XImage to a file "bitmap0.bmp" using the following code but when i open the newly created file with imageViewer in fedora it gives "Premature end of file" error and does not display anything. could anyone please point out the problem in the following code? typedef struct tagBITMAPFILEHEADER { WORD bfType; DWORD bfSize; WORD bfReserved1; WORD bfReserved2; DWORD bfOffBits; } BITMAPFILEHEADER; typedef struct tagBITMAPINFOHEADER{ DWORD biSize; LONG biWidth; LONG biHeight; WORD

fwrite() more than 2 GiB? [duplicate]

巧了我就是萌 提交于 2020-01-01 03:21:10
问题 This question already has answers here : Is fopen() limited by the filesystem? (4 answers) Closed 6 years ago . I have a set of files that I want to concatenate (each represents a part from a multi-part download). Each splitted file is about 250MiB in size, and I have a variable number of them. My concatenation logic is straight-forward: if (is_resource($handle = fopen($output, 'xb')) === true) { foreach ($parts as $part) { if (is_resource($part = fopen($part, 'rb')) === true) { while (feof(

Export a Json object to a text File

和自甴很熟 提交于 2019-12-31 13:49:12
问题 I'm trying to write a Json object (JsonExport) and I'd like to write its content into a text file. I'm using max4live to export data from Audio DAW to Json in order to export to a server, but after that I would like to see the whole Json Object in a text file: var txtFile = "test.txt"; var file = new File(txtFile); var str = JSON.stringify(JsonExport); file.open("write"); // open file with write access file.write(str); file.close(); The compiler runs with no error, but i can not get the text

Export a Json object to a text File

ε祈祈猫儿з 提交于 2019-12-31 13:48:22
问题 I'm trying to write a Json object (JsonExport) and I'd like to write its content into a text file. I'm using max4live to export data from Audio DAW to Json in order to export to a server, but after that I would like to see the whole Json Object in a text file: var txtFile = "test.txt"; var file = new File(txtFile); var str = JSON.stringify(JsonExport); file.open("write"); // open file with write access file.write(str); file.close(); The compiler runs with no error, but i can not get the text

fwrite in C giving different values in output files

ぐ巨炮叔叔 提交于 2019-12-31 04:21:38
问题 why are the output files different when I use fwrite in another function VERSUS fwrite in the same function? output1.txt contains garbage value like Ê, which is NOT correct output2.txt contains "b", which is correct #include <stdio.h> #include <string.h> void writeData(char *buf, char *path) { FILE *fp1; fp1 = fopen(path, "a"); fwrite(&buf, sizeof(char), strlen(buf), fp1); } int main () { char buf[2] = "a"; char buf2[3] = "yb"; char file1_path[12] = "output1.txt"; char file2_path[12] =

filemtime returns same value before and after modification of file

北慕城南 提交于 2019-12-31 00:58:11
问题 Im trying to get the last modified time of a file before and after i write to it using fwrite. But, i get the same values for some reason. <?php $i = filemtime('log.txt'); echo gmdate("h:i:s", $i); echo "<br/>"; $e=fopen('log.txt', 'w'); fwrite($e, "well well well"); $j = filemtime('log.txt'); echo gmdate("h:i:s", $j); ?> Now i modify 'log.txt' with a text editor about a minute before i run this script. So i should be getting about 40-60 seconds of time difference. If someone could point out

Why is fwrite writing more than I tell it to?

那年仲夏 提交于 2019-12-29 07:26:54
问题 FILE *out=fopen64("text.txt","w+"); unsigned int write; char *outbuf=new char[write]; //fill outbuf printf("%i\n",ftello64(out)); fwrite(outbuf,sizeof(char),write,out); printf("%i\n",write); printf("%i\n",ftello64(out)); output: 0 25755 25868 what is going on? write is set to 25755, and I tell fwrite to write that many bytes to a file, which is at the beginning, and then im at a position besides 25755? 回答1: If you are on a DOSish system (say, Windows) and the file is not opened in binary mode

How to check if a PHP stream resource is readable or writable?

做~自己de王妃 提交于 2019-12-29 05:46:07
问题 In PHP, how do I check if a stream resource (or file pointer, handle, or whatever you want to call them) is either readable or writable? For example, if you're faced with a situation where you know nothing about how the resource was opened or created, how do you check if it's readable? And how do you check if it's writable? Based on the testing that I've done (just with regular text files using PHP 5.3.3), fread() does not throw any errors at any level when the resource is not readable. It

C-fopen,fwrite,fread,fseek,fgets,popen,access笔记

丶灬走出姿态 提交于 2019-12-27 14:58:36
FILE * fopen(const char * path,const char * mode); 所需库: <stdio.h> 返回值 FILE是C语言定义的标准数据结构,如果open()失败,则返回NULL path 路径 mode 打开模式,包括有以下几种 r 以只读方式打开文件,该文件必须存在。 r+ 以读/写方式打开文件,该文件必须存在。 rb+ 以读/写方式打开一个二进制文件,只允许读/写数据。 rt+ 以读/写方式打开一个文本文件,允许读和写。 w 打开只写文件,若文件存在则长度清为 0,即该文件内容消失,若不存在则创建该文件。 w+ 打开可读/写文件,若文件存在则文件长度清为零,即该文件内容会消失。若文件不存在则建立该文件。 a 以附加的方式打开只写文件。若文件不存在,则会建立该文件,如果文件存在,写入的数据会被加到文件尾,即文件原先的内容会被保留(EOF 符保留)。 a+ 以附加方式打开可读/写的文件。若文件不存在,则会建立该文件,如果文件存在,则写入的数据会被加到文件尾后,即文件原先的内容会被保留(原来的 EOF 符不保留)。 wb 以只写方式打开或新建一个二进制文件,只允许写数据。 wb+ 以读/写方式打开或建立一个二进制文件,允许读和写。 wt+ 以读/写方式打开或建立一个文本文件,允许读写。 at+ 以读/写方式打开一个文本文件,允许读或在文本末追加数据