int main(int argc, char** argv) {
int file_size_limit = atoi(argv[1]);
char buf[BUFSIZ];
snprintf(buf, sizeof(buf), "find . -size +%dc -printf %M ", file_size_limit);
system(buf);
return 0;
}
The above code does not work and gives the warning: unknown conversion type character ‘M’ in format, how am I able to display only file permission with this structure. Thanks in advance!
I have tried to and double quotes on %M but this also gives an error, I cannot find a way to encode the %M deceleration into the find code since the whole command is passed in as an argument of the snprintf function
If you want to printf func to print a %, you need to prepend it with another %
snprintf(buf, sizeof(buf), "find . -size +%dc -printf %%M ", file_size_limit);
Also i suggest you to just printf to stdout the cmdline you just builded, you can then test the correctness of your code.
snprintf(buf, BUFSIZ, "find . -size +%dc -printf %%M ", file_size_limit);
fputs( buf, stdout );
Related
In pseudo code, I want put an arbitrary number of arguments to printf depending on the length of the argv, where the argv[1] is the format string.
int main(int argc, char *argv[]) {
printf(argv[1], argv[2], ...);
}
Uses can call the program as ./prog "%s %s" a b, ./prog "%s %s %s" a b c, and so on.
Could anybody let me know how to achieve this in C?
You need a loop for this:
int main(int argc, char *argv[])
{
int i;
for (i=1;i<argc;i++) {
printf("%s", argv[i]);
}
}
Here's something I just hacked together right now, it does does minimal parsing of the string and leaves most of it up to printf. It should also work with any number of arguments. Of course, since arguments are passed as char *s through the command line, this will only work with %s and its variants (and %%, but not sure if that counts as a format specifier).
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc, char *argv[])
{
if (argc < 2)
{
fprintf(stderr, "Usage: %s <format string>[ <args>]\n", argv[0]);
return 1;
}
// These pointers will constantly jump from format spec. to format spec.
char *last_fmt = argv[1], *next_fmt, *next_next_fmt;
char *buf = NULL; // a buffer to hold a substring of argv[1]
unsigned i = 2; // Used to iterate over argv[2+]
while (1)
{
next_fmt = last_fmt - 2;
do
{
if ((next_fmt = strchr(next_fmt + 2, '%')) == NULL)
{
/* Your compiler may warn about this line specifically (it did for me),
but rest assured that there are indeed no format specifiers
here really, printf is just needed for printing "%%"s as "%"s */
printf(last_fmt);
return 0;
}
} while (next_fmt[1] == '%');
next_next_fmt = next_fmt - 1;
do
{
if ((next_next_fmt = strchr(next_next_fmt + 2, '%')) == NULL)
{
printf(last_fmt == argv[1] ? last_fmt : next_fmt,
argv[i]);
return 0;
}
} while (next_next_fmt[1] == '%');
buf = malloc(next_next_fmt - last_fmt + 1);
memcpy(buf, last_fmt, next_next_fmt - last_fmt);
buf[next_next_fmt - last_fmt] = '\0';
printf(buf, argv[i]);
free(buf);
++i;
last_fmt = next_next_fmt;
}
}
An example of running:
./a.out "Hello %.2s World! %s" "foo" "bar"
Hello fo World! bar
./a.out "Hello %10s World!" "foo" "bar"
Hello foo World!
./a.out "Hello %5.2s World!" "random"
Hello ra World!
./a.out
Usage: ./a.out <format string>[ <args>]
./a.out "Hello %%s World %s" "a"
Hello %s World a
./a.out "%s %s %s" "a" "b" "c"
a b c
You could build upon this yourself, but if you want to handle other format specifiers, you'll have to do actual parsing of the string. At that point, you would basically be creating another printf.
You also might be a bit worried about the use of a not-string-literal passed to printf, but this is safe. There is guaranteed to be exactly 1 format specifier in each place I use printf (except in the first do loop, but there it is guaranteed to not have any arguments).
how to achieve this in C?
C language does not have reflection. You can't "dynamically create function calls" or inspect and then change your own source code. You have to know at compilation time how many arguments you are passing to a function. While it is simple to do printf("%s %s", "a", "b) inside C language, if you pass the same data to a program that was written in C language you have parse the data and write the logic yourself.
Such parser would take the string "%s %s" find the %s sequences and replace them for the string "a" and "b" and also print the space in between, etc. That parser has to be written in C and is basically a duplication of what printf program (not printf() C function) does. You may want to read some implementations of printf program: ex. coreutils printf.c or freebsd printf.c.
This isn't a great idea to begin with, it will be super-vulnerable to all manner of exploits, typos and bugs. But if you insist, you could do a dirty hack as follows:
Assuming the format string in argv[1] is %s %s %s, then each we can divide this string length by 3 to get the number of strings. Save for the final one, which isn't followed by a trailing space. So strlen(argv[1]) + 1 then divide by 3:
#define STR_N ((strlen(argv[1])+1)/3)
Next up we can take advantage of printf ignoring trailing arguments not corresponding to the format string. So we could do printf(argv[1], argv[2], argv[3]); just fine without actually passing that many arguments, long as the format string contains the correct amount of conversion specifiers. For example:
#define ARGV_LIST \
argv[2],\
argv[3],\
argv[4],\
argv[5],\
argv[6],\
argv[7],\
argv[8],\
argv[9]\
printf(argv[1], ARGV_LIST);
Then cook up something to convert the indices and make sure that array out of bounds never occurs:
#include <stdio.h>
#include <string.h>
#define STR_N ((strlen(argv[1])+1)/3)
#define INDEX(n) (STR_N>n? (n+2) : 0)
#define ARGV_LIST \
argv[INDEX(0)],\
argv[INDEX(1)],\
argv[INDEX(2)],\
argv[INDEX(3)],\
argv[INDEX(4)],\
argv[INDEX(5)],\
argv[INDEX(6)],\
argv[INDEX(7)],\
argv[INDEX(8)],\
argv[INDEX(9)]\
int main(int argc, char *argv[])
{
printf(argv[1], ARGV_LIST);
return 0;
}
Tested in Windows with prog.exe "%s %s %s %s %s" hello this is a test gives output:
hello this is a test
I have a binary file which prints the result instead of returning the value, if I execute it using cmd I am getting printed text, I managed to execute it from C code but it seems like I can not get the text it usually prints to be stored in a variable I can use later for further decisions.
I do not have that much of experience in C and I googled a lot.
I came across the idea of using clip but my cmd is saying that clip command can not be found.
any help or ideas would be appreciated.
The correct function pair to use on POSIX systems is popen() and
pclose(). You can perhaps use Microsoft's _popen() and
_pclose() unless the warning 'This API cannot be used in applications that execute in the Windows Runtime' matters to you.
You would use it more or less like this. I've had to invent the name of the command you wish to execute since the question doesn't specify that. I chose ./example.exe as the name — and I'm assuming it needs no arguments.
char cmd[] = "./example.exe";
FILE *fp = popen(cmd, "r");
if (fp != NULL)
{
char buffer[4096];
size_t nbytes;
while ((nbytes = fread(buffer, sizeof(buffer), sizeof(char), fp)) != 0)
{
…process nbytes of data…
…it is not a null-terminated string unless you add the null byte…
}
pclose(fp);
}
else
{
…report error for failure to execute command…
}
You can use the system function from <stdlib.h> to run the command you want. To get the command's output, you modify your command like in this question to save the command's output to a file. Then you can use the file I/O functions in <stdio.h> to process the command output.
In Linux, you may do command substitution and pass its result as arguments to the program, Something like this
./your_program "$(/path/to/your/binary/file)"
Suppose your main is
int main(int argc,char* argv[]){
.
.
return 0;
}
Acess the arguments like argv[1] and so.
Here the $(command) does the substitution and it passes the printed values from the binary as arguments to the pgm. Hope this helps.
Use snprintf function. For e.g.
snprintf(cmdbuff, BUFFER_LEN, "dmidecode --type 17 | grep -i Size | grep -o '\\<[0-9]*\\>' | paste -sd+ | bc");
Here cmdbuff is character array where command will be stored , BUFFER_LEN is a size of the character array
Then use popen and fgets to get the output of command into some buffer as shown below
if((fd = popen(cmdbuff,"r")) != NULL)
{
fgets(buffer, BUFFER_LEN, fd);
sprintf(vnfc_configured_memory, "%s", buffer);
vnfc_configured_totalRAM = atof(vnfc_configured_memory);
}
I'm trying to create a simple script on my server, basically I would like to sent a string and display it via system function...
#include <stdio.h>
int main()
{
char txt[100];
printf("Insert a text: ");
fgets(txt, 100, stdin);
system("echo %s"), txt;
return 0;
}
Rght now I'm not getting any string just "%s"
any idea why?
system("echo %s"), txt;
This isn't doing what you think; it's an expression which evaluates to txt. Since evaluating txt has no side effects, and since you're not capturing the result of the expression anywhere, adding , txt after the system call essentially does nothing. See this question for some information on the "comma"-operator in C.
Moreover, system doesn't support the use of printf-style format specifiers, so the %s in your string literal doesn't have any special meaning; it's just going to be echoed exactly as written, as you've seen. If you want to construct a command at runtime for use with system, you will have to do so with sprintf or similar.
The prototype to system() is:
int system(const char * command);
From man 3 system:
executes the shell command specified in command
From this we can safely assume s refers to a C-"string".
So prepare the string using for example snprintf():
char s[1024];
snprintf(s, 1024 -1, "echo %s", txt); /* -1 for the C-"string"'s 0-terminator */
Then pass it:
system(s);
Instead of system("echo %s"), txt; try this:
printf("%s", txt);
the system statement will not format the output, like printf.
suggest using:
#include <stdio.h>
#include <stdlib.h> // system()
#include <string.h> // strcpy(), strcat()
#define BUF_LEN (100)
int main()
{
char output[10+BUF_LEN] = "echo ";
char txt[BUF_LEN] = {'\0'};
printf("Insert a text: ");
fgets(txt, BUF_LEN, stdin);
strcat( output, txt );
system( output );
return 0;
}
The above code works very nicely, however;
do not include any command separators, semicolons, or other characters that would be interpreted by the shell in the input string.
is it possible to save information gather from a sprintf into a variable? The lines of code below are an example to better illustrate my question.
char fileName;
fileName = sprintf(command, "find -inum %i -type f", iNode);
The purpose is to find the file name associated with the inode number, then run "stat" on that file name.
I think you want something like this:
FILE *fp;
char cmd[1024];
char filename[1024];
sprintf(cmd, "find -inum %i -type f", iNode);
fp = popen(cmd);
fgets(filename, sizeof filename, fp);
pclose(fp);
At the end of this code, filename will contain the fist line produced by the cmd.
I'm new to C and I'd like to ask about running a C program and supplying input at the same time.
What I would like to do is run a program (ex. fileOpener) and also state which file to open
./fileOpener < filename1
I've tried it already and it works fine, but what do I use to know what filename1 is? That way I can open the file with
fp = fopen(filename1, "r")
Thanks.
Edit: OK, I'll try to explain a bit more. If there wasn't a "<" then I could just use command line arguments as I have done before, but when I tried it with the <, it didn't work
Specifically: fileOpener code:
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv[]){
printf("%s", argv[1]);
}
when I use ./fileOpener < filename1 the output is ./fileOpener
I used gcc -o fileOpener fileOpener.c as the compiler
int main(int argc, char *argv[])
You can name them whatever you want, but these are the normal names.
argc is non-negative. It gives the number of useful elements in argv.
If argc is positive, argv[0] contains the program name. Then argv[1] through argv[argc - 1] point to character arrays that contain the program's command line arguments.
For example, if I run a program at the command line, such as
unzip filename.zip
argc will equal 2; and argv[0] will compare equal to "unzip"; and argv[1] will compare equal to "filename.zip".
Source
You can't do that, if you use redirection (i.e. "< filename") the file is opened by the system. You could discover the name, but it's non-portable, and anyway useless since the file is already open. Just use stdin instead of fp, and you need not use fopen() (nor fclose()):
int main()
{
char buffer[1024];
// fgets() reads at most 1024 characters unless it hits a newline first
// STDIN has been already opened by the system, and assigned to data flowing
// in from our file ( < inputFile ).
fgets(buffer, 1024, stdin);
printf("The first line of input was: %s", buffer);
}
A different approach is to use arguments:
int main(int argc, char **argv)
{
FILE *fp = NULL;
char buffer[1024];
if (argc != 2)
{
fprintf(stderr, "You need to specify one argument, and only one\n");
fprintf(stderr, "Example: %s filename\n", argv[0]);
// Except that argv[0], this program's name, counts.
// So 1 argument in command line means argc = 2.
return -1;
}
printf("I am %s. You wanted to open %s\n", argv[0], argv[1]);
fp = fopen(argv[1], "r");
fgets(buffer, 1024, stdin);
printf("The first line of input was: %s", buffer);
fclose(fp); fp = NULL; // paranoid check
return 0;
}
You need setup your program to take a command line argument. Here's a good tutorial that solves your exact question:
http://www.cprogramming.com/tutorial/c/lesson14.html
A program's main function in C has two arguments:
int main(int nArgs, char *pszArgs[]) {}
That first argument tells the program how many parameters were passed onto the program when you ran it. Usually, this will just be 1, because it includes the program's name.
The second argument is a table of strings, which can be accessed thus (the program below prints the parameters given to it):
int main(int nArgs, char *pszArgs[])
{
int i = 0;
while (i < nArgs)
{
printf("param %d: %s\n", i, pszArgs[i]);
i++;
}
return 0;
}