How to output single character in C with write? - c

Could someone instruct me on how to print a single letter, for example "b" in C while using only the write function (not printf).
I'm pretty sure it uses
#include <unistd.h>
Could you also tell me how the write properties work? I don't really understand
int write( int handle, void *buffer, int nbyte );
Could some of you guys toss in a few C beginner tips as well?
I am using UNIX.

You have found your function, all you need now is to pass it proper parameters:
int handle = open("myfile.bin", O_WRONLY);
//... You need to check the result here
int count = write(handle, "b", 1); // Pass a single character
if (count == 1) {
printf("Success!");
}
I did indeed want to use stdout. How do I write a version to display the whole alphabet?
You could use a pre-defined constant for stdout. It is called STDOUT_FILENO.
If you would like to write out the whole alphabet, you could do it like this:
for (char c = 'A' ; c <= 'Z' ; c++) {
write(STDOUT_FILENO, &c, 1);
}
Demo.

Let's see the man page of write(), which says,
ssize_t write(int fd, const void *buf, size_t count);
Description
write() writes up to count bytes from the buffer pointed buf to the file referred to by the file descriptor fd.
As per your requirement, you need to pass an address of a buffer containing b to print to standard output.
Let's see some code along with, shall we?
#include <stdio.h>
#include <unistd.h>
int main(void) {
char b = 'b';
write(STDOUT_FILENO, &b, 1);
return 0;
}
Let me explain. Here, the STDOUT_FILENO is the file descriptior for standard output as defined in unistd.h, &b is the address of the buffer containing 'b' and the number of bytes is 1.

Also valid is:
#include <stdio.h>
#include <unistd.h>
int main(void) {
char b[1] = {'b'};
write(STDOUT_FILENO, b, 1);
return 0;
}
Or:
#include <stdio.h>
#include <unistd.h>
int main(void) {
char b[1] = "b";
write(STDOUT_FILENO, b, 1);
return 0;
}

Related

(C) generating random data using /dev/urandom and writing it into file

I'm pretty new to C, and I'm not sure what is wrong with this piece of code I have written. It is supposed to open (or create if needed) a file using a name specified and in that file write a bunch of signs taken from the /dev/urandom file. I need a precise amount of elements each one of a specified length. I need to be able to later edit that created file, but I wanted to focus on creating this generator part first. It doesn't show any errors when compiling or when I try to execute it, but still nothing happens - the file still doesn't exist anywhere. What's wrong with it?
int main(){
void generate(char str[], int a, int b);
}
void generate(char str[], int a, int b){
int n=0;
char fname[128];
strcpy(fname,str);
strcpy(fname, ".txt");
FILE *myFile = fopen(fname, "w");
FILE *randomiser = fopen("/dev/urandom", "r");
char bufor[a];
size_t result = fread (bufor, b, a, randomiser);
size_t end = fwrite (bufor, b, a, myFile);
fclose(myFile);
fclose(randomiser);
}
#edit
change the tile as someone suggested and changed a code a bit since im still trying to work it out
I forgot to mention that the whole point of this function is for it to be called in terminal as for example ./program generate data 100 100.
Here is the exact solution you asked for
There you go. I use open instead of fopen, read a certain amount of char out of /dev/urandom then write it to fd2 (the file created),
Be carefull the size of the buffer, be careful The NULL (\0) terminate the string
#include <sys/types.h>
#include <sys/uio.h>
#include <unistd.h>
#include <fcntl.h>
#include <stdlib.h>
int generate(char **av)
{
int fd1 = open("/dev/urandom", O_RDONLY);
int fd2 = open(av[1], O_CREAT | O_RDWR, S_IRWXU);
int size = atoi(av[2]);
int row = atoi(av[3]);
int i = -1;
int j = -1;
if (fd1 == -1 || fd2 == -1 || size <= 0 || row <= 0 )
return (1);
char buf[size];
while (++i < row-1)
{
read(fd1, buf, size);
buf[size -1] = '\0';
write(fd2, buf, size -1);
while (++j < size)
buf[j] = '\0';
}
close(fd1);
close(fd2);
return (0);
}
int main(int ac, char **av)
{
int s;
if (ac < 4)
return (1);
generate(av);
return (0);
}
for 10 * 100 char into file 'blabla' Use with :
$~> gcc prog.c -o program
$~> ./program blabla 100 10
Will create / open the file "blabla" read 100 from /dev/uramdom and wirte to blabla
To explain what you did wrong here look :
This is a declaration of a function
void generate(int i, char *a);
This is a call to a function
generate(25, "Blablabla");
sometime when you declare a function you also instanciate it aswell
int generate(int i, char *a)
{
Operation;
operation;
return (0);
}
Declaration, and instanciation are different, let's assume i wrote funtion generate under function main in the above code example, function main would not know what is "generate" but if you declare it before main just like this it works
void generate(int i, char *a);
Here's what I made out of your source code with minimal changes
#include <stdio.h>
void generate(char basename[], size_t size);
int main(void) {
generate("foobar", 42);
}
void generate(char basename[], size_t size) {
char fname[128];
sprintf(fname, "%s.txt", basename);
FILE *myFile = fopen(fname, "w");
FILE *randomiser = fopen("/dev/urandom", "r");
char bufor[size];
size_t result = fread(bufor, 1, size, randomiser);
fwrite(bufor, 1, result, myFile);
fclose(myFile);
fclose(randomiser);
}
The problem is that your program never calls the generate() function, as #fredrik pointed out in comments. This:
void generate(char str[], int a, int b);
is a function declaration, not a call. Although it is legal to put such a declaration inside a function, it is much more conventional to put it outside and preceding the function(s) that will be doing the calling. And in this case, you could avoid using a separate declaration at all by just moving the main() function to the end, after the definition of generate().
But that still doesn't get around the fact that you need to call the function if you want it to do anything. And to call it, you will need to provide arguments. For example,
generate("random_data", 10, 4);
From your description, you'll want to use arguments determined from evaluation of command-line parameters, but that aspect would be a separate question.

Segmentation fault using strcmp and char *arr[]

I am working on creating a shell and I haven't use C for a while. I have the shell initizing properly but when I try to compare the user input to an array of strings I have I get a segmentation fault. I was planning on adding casce statements in a the for loop to initiate each of the processes once they are called by the user. I haven't included those since I have been trying to figure out how to get the user input to match with a value in my string array. Under debug I was only receiving the first character of the builtins[j] value which kind of makes since since it is a pointer right. However I am stuck and could use some ideas for why this isn't returning 0 when I input "exit". Thanks
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <sys/stat.h>
#include <fcntl.h>
//This code is for creating a basic shell
void init_shell(int num, char *prompt[]){
char s1[] = "-p";
int result;
if(num>1){
result = strcmp(s1, prompt[1]);
if(result==0){
printf("%s>$", prompt[2]);
}else{
printf("308sh>$");
}
}
//printf("%s\n %s\n %s\n %d\n", prompt[0], prompt[1], prompt[2], result);
else{
printf("308sh>$");
}
}
//The infinite loop for accepting user input until it closes
int main(int argc, char *argv[]){
const char *builtins[7];
builtins[0] = "exit\n";
builtins[1] = "pid\n";
builtins[2] = "ppid\n";
builtins[3] = "cd\n";
builtins[4] = "pwd\n";
builtins[5] = "set\n";
builtins[6] = "get\n";
char usr_in[]="";
char cmp[]="";
while(1==1){
init_shell(argc, argv);//intial prompt for the shell
fgets(usr_in,100,stdin);
//Check for builtin Commands
int cmds_size = 7;
int j=0;
int res;
for(j; j<cmds_size; j++){
res=strcmp(usr_in, hold);
if(res==0){
printf("Execucting\n");
}
else{
printf("no command\n");
}
}
}
return(0);
}
The issue here is that you're writing the user's input to a buffer that isn't big enough to hold anything other than a null terminator.
char user_in[] = "";
The above line tells the C compiler that you need just enough space to store [ '\0' ], which is a single byte. The C compiler doesn't know that you may later write a 100-byte string to that buffer.
When you write to the buffer, the user's input overflows and will overwrite other values in your stack. Since the other values in your stack are pointers, what'll happen is you'll run into seg-faults, since you're writing character values into those bytes, but interpreting them as char pointers.
You are correctly limiting the size of the allowed input from the user to 100 characters, but you should make sure that your buffer is big enough to hold the value you're reading in:
char user_in[101];
for(int i = 0; i < sizeof(user_in) / sizeof(user_in[0]); i++) {
user_in[i] = 0; // Since this is allocated on the stack *in main*, this
// shouldn't be necessary
}
Here's one example of how you can rewrite your main method:
#include <stdio.h>
#include <string.h>
typedef enum { false, true } bool; // If you don't have this
// defined already
int main(int argc, char *argv[]) {
const char *builtins[7];
builtins[0] = "exit\n";
builtins[1] = "pid\n";
builtins[2] = "ppid\n";
builtins[3] = "cd\n";
builtins[4] = "pwd\n";
builtins[5] = "set\n";
builtins[6] = "get\n";
char user_in[101];
for(int i = 0; i < sizeof(user_in) / sizeof(user_in[0]); i++) {
user_in[i] = 0;
}
while(1) {
printf("Enter a command: ");
fgets(user_in, 100, stdin);
bool found = false;
for(int i = 0; i < sizeof(builtins) / sizeof(builtins[0]); i++) {
if (!strcmp(user_in, builtins[i])) {
printf("Found command %s", builtins[i]);
found = true;
break;
}
}
if (!found) {
printf("Didn't find command\n");
}
}
return 0;
}
Also, regarding your function init_shell: you're checking to see if argc is greater than 1, but that only guarantees that argv[1] is defined; it doesn't guarantee that argv[2] is defined. (Remember, argc is the size of the argv array, where the first element is the name of the program being executed). You want to make sure that argc is at least 3 before checking for the prompt flag in the way you are.
It may be overkill for your use-case, but consider using the getopt function for getting a custom prompt value from the user. See http://man7.org/linux/man-pages/man3/getopt.3.html for documentation regarding that method.

C, perform file operations with array [duplicate]

Is there any way to create a memory buffer as a FILE*. In TiXml it can print the xml to a FILE* but i cant seem to make it print to a memory buffer.
There is a POSIX way to use memory as a FILE descriptor: fmemopen or open_memstream, depending on the semantics you want: Difference between fmemopen and open_memstream
I guess the proper answer is that by Kevin. But here is a hack to do it with FILE *. Note that if the buffer size (here 100000) is too small then you lose data, as it is written out when the buffer is flushed. Also, if the program calls fflush() you lose the data.
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char **argv)
{
FILE *f = fopen("/dev/null", "w");
int i;
int written = 0;
char *buf = malloc(100000);
setbuffer(f, buf, 100000);
for (i = 0; i < 1000; i++)
{
written += fprintf(f, "Number %d\n", i);
}
for (i = 0; i < written; i++) {
printf("%c", buf[i]);
}
}
fmemopen can create FILE from buffer, does it make any sense to you?
I wrote a simple example how i would create an in-memory FILE:
#include <unistd.h>
#include <stdio.h>
int main(){
int p[2]; pipe(p); FILE *f = fdopen( p[1], "w" );
if( !fork() ){
fprintf( f, "working" );
return 0;
}
fclose(f); close(p[1]);
char buff[100]; int len;
while( (len=read(p[0], buff, 100))>0 )
printf(" from child: '%*s'", len, buff );
puts("");
}
C++ basic_streambuf inheritance
In C++, you should avoid FILE* if you can.
Using only the C++ stdlib, it is possible to make a single interface that transparently uses file or memory IO.
This uses techniques mentioned at: Setting the internal buffer used by a standard stream (pubsetbuf)
#include <cassert>
#include <cstring>
#include <fstream>
#include <iostream>
#include <ostream>
#include <sstream>
/* This can write either to files or memory. */
void write(std::ostream& os) {
os << "abc";
}
template <typename char_type>
struct ostreambuf : public std::basic_streambuf<char_type, std::char_traits<char_type> > {
ostreambuf(char_type* buffer, std::streamsize bufferLength) {
this->setp(buffer, buffer + bufferLength);
}
};
int main() {
/* To memory, in our own externally supplied buffer. */
{
char c[3];
ostreambuf<char> buf(c, sizeof(c));
std::ostream s(&buf);
write(s);
assert(memcmp(c, "abc", sizeof(c)) == 0);
}
/* To memory, but in a hidden buffer. */
{
std::stringstream s;
write(s);
assert(s.str() == "abc");
}
/* To file. */
{
std::ofstream s("a.tmp");
write(s);
s.close();
}
/* I think this is implementation defined.
* pusetbuf calls basic_filebuf::setbuf(). */
{
char c[3];
std::ofstream s;
s.rdbuf()->pubsetbuf(c, sizeof c);
write(s);
s.close();
//assert(memcmp(c, "abc", sizeof(c)) == 0);
}
}
Unfortunately, it does not seem possible to interchange FILE* and fstream: Getting a FILE* from a std::fstream
You could use the CStr method of TiXMLPrinter which the documentation states:
The TiXmlPrinter is useful when you
need to:
Print to memory (especially in non-STL mode)
Control formatting (line endings, etc.)
https://github.com/Snaipe/fmem is a wrapper for different platform/version specific implementations of memory streams
It tries in sequence the following implementations:
open_memstream.
fopencookie, with growing dynamic buffer.
funopen, with growing dynamic buffer.
WinAPI temporary memory-backed file.
When no other mean is available, fmem falls back to tmpfile()

Is there a way to print something in case of a segfault? C

For example, If my program segaults, instead of gcc printing to the console "Segmentation Fault" can I have it print "Ya dun goofed"?
Segfaults are generally caused by dereferencing a garbage pointer. Therefore, while the literal answer to what you asked is that, as kaylum said, you can catch SIGSEGV in a signal handler, the better answer is that, before you use a pointer, you should ask yourself, “How do I know that this pointer is valid and that I am staying within the bounds of my array?"
If you don’t know that, your program has a bug. If you think you do, you can turn the assumption into an assertion which, since your pointer is valid, will always pass. For example:
void fill_array( unsigned fill_this_many,
size_t array_size,
int a[array_size] )
{
assert(a);
assert( array_size >= fill_this_many );
for ( unsigned i = 0; i < fill_this_many; ++i )
a[i] = f(i);
return;
}
You’ll now get a detailed message when you’re about to dereference a null pointer or write past the end of your array, which will contain more useful information for debugging than, "There was a segfault somewhere," and it might even save you from silent memory corruption too.
If you want to write your own message, you can define a wrapper such as:
#include <stdio.h>
#include <stdlib.h>
void fatal_error_helper( const char* file, int line, const char* restrict message )
{
fflush(stdout); // Don’t cross the streams!
fprintf( stderr, "\nError in %s, line %d: %s\n", file, line, message );
exit(EXIT_FAILURE);
}
#define fatal_error(message) fatal_error_helper( __FILE__, __LINE__, (message) )
int main(void)
{
int *big_array = calloc( 1073741824UL, sizeof(int) );
if (!big_array)
fatal_error("Not enough memory.");
return EXIT_SUCCESS;
}
And a contrived example of how to do bounds-checking at compile time, so as to fail gracefully if your constants change:
#include <assert.h>
#include <ctype.h>
#include <stdio.h>
#include <stdlib.h>
#define LENGTH 14U
#define M 5U
int main(void)
{
char message[LENGTH] = "hello, world!";
static_assert( M < LENGTH, "Tried to capitalize more letters than the array can hold." );
for ( unsigned i = 0; i < M; ++i )
message[i] = toupper(message[i]);
printf( "%s\n", message );
return EXIT_SUCCESS;
}
#include <signal.h>
#include <stdio.h>
#include <unistd.h>
#include <string.h>
#include <stdlib.h>
void segv_handler(int sig)
{
(void)sig;
const char *msg = "Hello signal handler!";
size_t len = strlen(msg);
write(STDERR_FILENO, msg, len);
abort();
}
int main()
{
struct sigaction act;
act.sa_handler = segv_handler;
sigemptyset(&act.sa_mask);
act.sa_flags = 0;
sigaction(SIGSEGV, &act, NULL);
int *nullint = 0;
*nullint = 4;
return 0;
}
EDIT: I tough code is pretty much explanation how to do it. Of course there is a lot details that needs to be taken into account when writing signal handlers.
Basic limitation is that signal handler can't access any variable/structure that isn't written to atomically because handler could be called between any two instructions in your program. That means no calls to heap memory management, buffered io like printf, etc.
More details what the code does can be found from man pages stdout, sigaction and write.

c - cheak readable bytes

need little help form you guys
i just wanna to cheak if the byte is readable or not, i have search for sulution but not find
hope you will help me
i have this code i need if tag that cheak if byte is readable
#include <windows.h>
#include <iostream>
#include <cstdlib>
#include <stdio.h>
void main()
{
float ramsize;
char *ch;
unsigned int j=128,readbyte;
long i;
MEMORYSTATUSEX statex;
statex.dwLength = sizeof (statex);
GlobalMemoryStatusEx (&statex);
ramsize = statex.ullTotalPhys;
for(i=0;i<ramsize;i = i+1)
{
ch = (char*) i;
readbyte = *ch;
// if readbyte is readable
printf("you have readable byte in address: %x , that contain in Binary:",&readbyte);
for(i=0;i<8;i++)
{
if(readbyte&j)
printf("1");
else
printf("0");
j=j>>1;
}
putchar('\n');
// if readbyte is not readable
printf("Sorry: you cant read this byte: %x",&readbyte);
}
}
If a byte is not readable the OS will send a signal to your process. You need to catch that signal or your program will terminate.
Read up on signals in your course textbook.

Resources