I want to open a file using the arguments when executing it, for example:
./Project 123.txt
I can make this work to some extent, but if I try to pass the arguments into a function, and then call the function, something is not working quite right.
Here is the function I have, which will read the lines from a text file:
int numeroLinhas(){
int ch;
FILE *fp;
int linhas=0;
fp = fopen("123.txt","r");
while( ( ch = fgetc(fp) ) != EOF ){
if(ch=='\n'){
linhas++;
}
}
fclose(fp);
fprintf(stats, "linhas: %d\n", linhas);
}
int main(int argc, char *argv[]){
FILE *fp;
fp = fopen(argv[1],"r");
numeroLinhas();
}
So, I would like to know how can I pass the argv[] has an argument into my numeroLinhas() function, so that I don't have to call the name of the file at all during the coding, just when executing.
The problem is that you have hard-coded the file name to open. Modify your function to take an argument, a string naming the file.
Then just pass the correct argv entry as the argument to the function, after checking the number of arguments to the program first.
Pass the name of the file to the function that counts the lines in the file:
#include <stdio.h>
void numeroLinhas(const char *filename)
{
int ch;
int linhas = 0;
FILE *fp = fopen(filename, "r");
if (fp == 0)
{
fprintf(stderr, "Failed to open file %s for reading\n", filename);
return;
}
while ((ch = fgetc(fp)) != EOF)
{
if (ch == '\n')
linhas++;
}
fclose(fp);
fprintf(stats, "linhas: %d no %s\n", linhas, filename);
}
int main (int argc, char *argv[])
{
for (int i = 1; i < argc; i++)
numeroLinhas(argv[i]);
return 0;
}
Error check fopen() calls; they fail. Handle the int returned by fgetc() correctly. Change the function to return void since there isn't a return. Report errors on standard error. Include file name in the outputs (error and routine). Invoke the function on all arguments provided. Don't say anything if no arguments are provided — or add if (argc < 2) { fprintf(stderr, "Usage: %s file [...]\n", argv[0]); return 1; } before the loop in main().
Sorry about mixed English/Portugese messages.
int numeroLinhas(char *filename){
char ch;
FILE *fp;
int linhas=0;
fp = fopen(filename,"r");
while( ( ch = fgetc(fp) ) != EOF ){
if(ch=='\n'){
linhas++;
}
}
fclose(fp);
fprintf(stats, "linhas: %d\n", linhas);
}
int main (int argc, char *argv[]){
//should have an if for argc size
char *filename = argv[1];
numeroLinhas(filename);
}
Related
I am trying to read a file in C. First I am calculating the lines in the file:
int main(int argc, char* argv[])
{
if (argc < 2)
{
printf("No file specified");
exit(1);
}
FILE* pFile;
char currentCharacter;
int lines = 1;
pFile = fopen(argv[1], "r");
for (currentCharacter = getc(pFile); currentCharacter != EOF; currentCharacter = getc(pFile))
{
if (currentCharacter == '\n') lines++;
}
...
}
After calculating the lines in the file, I tried reading one by one, like this:
char currentLine[255];
for (int i = 1; i <= lines; i++)
{
fgets(currentLine, 255, pFile);
printf("%s\n", currentLine);
}
fclose(pFile);
But everytime I run it, I am getting this output:
²a
When I try to remove the for loop and place fgets() and printf() outside, it prints NOTHING
If you are wondering, here is the content of the file I am trying to read:
test.txt
test1
test2
test3
NOTE: The file is being successfully opened as it is counting the lines correctly.
As said in the comments, no need to count the lines. Just stop when there is nothing more to read. That is, when fgets returns NULL.
#include<stdio.h>
#include<stdlib.h>
int main(int argc, char* argv[])
{
if (argc < 2)
{
printf("No file specified");
exit(1);
}
FILE* pFile = fopen(argv[1], "r");
if(pFile==NULL)
{
printf("File is not found");
exit(1);
}
char currentLine[256];
while(fgets(currentLine, 256, pFile))
{
printf("%s", currentLine);
}
return 0;
}
I'm writing a program said in this post title. I take reference at this webpage.
https://www.includehelp.com/c-programs/c-program-to-print-given-number-of-lines-of-a-file-like-head-command-in-linux.aspx
Here are the codes from that webpage.
#include <stdio.h>
int main(int argc, char * argv[])
{
FILE *fp; // file pointer
char *line = NULL;
int len = 0;
int cnt = 0;
if( argc < 3)
{
printf("Insufficient Arguments!!!\n");
printf("Please use \"program-name file-name N\" format.\n");
return -1;
}
// open file
fp = fopen(argv[1],"r");
// checking for file is exist or not
if( fp == NULL )
{
printf("\n%s file can not be opened !!!\n",argv[1]);
return 1;
}
// read lines from file one by one
while (getline(&line, &len, fp) != -1)
{
cnt++;
if ( cnt > atoi(argv[2]) )
break;
printf("%s",line); fflush(stdout);
}
// close file
fclose(fp);
return 0;
}
My problem is the getline function. Since I'm not using Linux that function's giving error in my compiler. I tried to change it to fgets function. This is my revised codes.
I got two errors in the line ' while (fgets(&line, bufferLength, fp) != -1)'.
Error: passing argument 1 of 'fgets' from incompatible pointer type.
Error: comparison between pointer and integer.
My question is - how can I modify the program using fgets? Many thanks to anyone who can work this out.
#include <stdlib.h>
#include <stdio.h>
int main(int argc, char *argv[])
{
FILE *fp; // file pointer
char *line = NULL;
int bufferLength = 255;
int cnt = 0;
if( argc < 3)
{
printf("Insufficient Arguments!!!\n");
printf("Please use \"program-name file-name N\" format.\n");
return -1;
}
// open file
fp = fopen(argv[1],"r");
// checking for file is exist or not
if( fp == NULL )
{
printf("\n%s file can not be opened !!!\n",argv[1]);
return 1;
}
// read lines from file one by one
while (fgets(&line, bufferLength, fp) != -1)
{
cnt++;
if ( cnt > atoi(argv[2]) )
break;
printf("%s",line);
fflush(stdout);
}
// close file
fclose(fp);
return 0;
}
Your program should compile and run correctly follows:
//c program to print given number of lines from beginning of a file
//file name and number of lines must be supply as command line argument
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char * argv[])
{
FILE* fp; // file pointer
char* line = malloc(255);
int bufferLength = 255;
int cnt = 0;
if( argc < 3)
{
printf("Insufficient Arguments!!!\n");
printf("Please use \"program-name file-name N\" format.\n");
return -1;
}
// open file
fp = fopen(argv[1],"r");
// checking for file is exist or not
if( fp == NULL )
{
printf("\n%s file can not be opened !!!\n",argv[1]);
return 1;
}
// read lines from file one by one
while (fgets(line,bufferLength, fp))
{
cnt++;
if ( cnt > atoi(argv[2]) )
break;
printf("%s",line);
fflush(stdout);
}
// close file
fclose(fp);
free(line);
return 0;
}
we have two main problems, first
char * line = NULL;
line is a line of characters, a string if you want to call it that, so we must reserve enough memory to accommodate a complete line, and we do this with the malloc function, as seen in the program, the other problem we have with fgets, this function returns a pointer therefore we cannot compare the value returned by fgets with an integer, the declaration
while (fgets (line, bufferLength, fp))
is equivalent to running the loop while fgets is other than NULL. Finally we must use line instead of &line, the latter asks for the address of the line pointer, and not the address it points to.
There's no need to keep track of more than a single character. Reading full lines is overkill. Just do:
#include <stdio.h>
#include <stdlib.h>
FILE *
xfopen(const char *path, const char *mode)
{
FILE *fp = fopen(path, mode);
if( fp == NULL ) {
perror(path);
exit(EXIT_FAILURE);
}
return fp;
}
int
main(int argc, char **argv)
{
int count = argc > 1 ? strtol(argv[1], NULL, 10) : 1;
FILE *in = argc > 2 ? xfopen(argv[2], "r") : stdin;
int line = 0;
int c;
while( line < count && ( c = fgetc(in)) != EOF ) {
putchar(c);
if( c == '\n' ) {
line += 1;
}
}
}
Note that I've reversed the order of the arguments, so that stdin is read if only a count is given.
This question already has answers here:
Read from file or stdin
(6 answers)
Closed 2 years ago.
I have a program that calculates a lottery tickets (this tickets are in a file.txt), and writes the winners tickets in another file. I have a subfunction called evaluate_tickets(file, lottery_numers, winner....)
In shell I write: ./program arg1 arg2... (arg1, arg2 are text files i.e. file.txt)
But now, I want to do ./program < file.txt. The problem is that I don't know how to send the parameter "file" of evaluate_tickets because I receive information by stdin.
Define a stream pointer FILE *fp; to read to input file:
If you want the input to be read from a file, use fp = fopen(filename, "r"); to open the file and close the stream after processing with fclose(fp);.
If you want the input to be read from standard input, just assign fp = stdin; instead of using fopen().
Here is a short example:
#include <stdio.h>
int main(int argc, char *argv[]) {
FILE *fp;
int c, lines;
if (argc > 1) {
fp = fopen(argv[1], "r");
if (fp == NULL) {
fprintf(stderr, "cannot open %s\n", argv[1]);
return 1;
}
} else {
fp = stdin; /* read from standard input if no argument on the command line */
}
lines = 0;
while ((c = getc(fp)) != EOF) {
lines += (c == '\n');
}
printf("%d lines\n", lines);
if (argc > 1) {
fclose(fp);
}
return 0;
}
Here is the same example with a cleaner approach, passing stdin or an open FILE pointer to an ad hoc function. Note how it handles all command line arguments:
#include <stdio.h>
void count_lines(FILE *fp, const char *name) {
int c, lines = 0;
while ((c = getc(fp)) != EOF) {
lines += (c == '\n');
}
printf("%s: %d lines\n", name, lines);
}
int main(int argc, char *argv[]) {
FILE *fp;
if (argc > 1) {
for (int i = 1; i < argc; i++) {
fp = fopen(argv[i], "r");
if (fp == NULL) {
fprintf(stderr, "cannot open %s\n", argv[i]);
return 1;
}
count_lines(fp, argv[i]);
fclose(fp);
}
} else {
/* read from standard input if no argument on the command line */
count_lines(stdin, "<stdin>");
}
return 0;
}
I am trying to figure out why this is not printing, I am trying to print each letter from a text file that is inputted through command prompt, but I am just getting an empty output... What am I doing wrong, and why does this not work? I feel like this logically should work. Thanks.
int main(int argc, char *argv[]) {
FILE *fp;
int i;
for (i = 1; i < argc; i++) {
printf("%s\n", argv[i]);
fp = fopen(argv[i], "r");
while (!feof(fp)) {
puts(fp);
}
fclose(fp);
}
return 0;
}
You are attempting to print a file pointer:
puts(fp);
Read the manual of puts() -that's not what it takes.
To read char-by-char and print on the stdout, you can do:
int ch;
fp = fopen(argv[i], "r");
if (!fp) {
perror("fopen");
exit(1);
}
while((ch=fgetc(fp)) != EOF) {
putchar(ch);
}
flcose(fp);
Unless you are passing multiple file names as arguments, your outer loop doesn't make much sense.
Your program has multiple problems:
You do not test the return value of fopen(): the program invokes undefined behavior if any of the command line arguments cannot be opened as a stream for reading.
while(!feof(fp)) is incorrect. Read this: Why is “while ( !feof (file) )” always wrong?
puts(fp); is incorrect as fp is a FILE *, not a string. Use a loop to copy the file contents one byte at a time.
Here is a corrected version:
#include <errno.h>
#include <stdio.h>
#include <string.h>
int main(int argc, char *argv[]) {
FILE *fp;
int i, c;
for (i = 1; i < argc; i++) {
fp = fopen(argv[i], "r");
if (fp == NULL) {
fprintf(stderr, "cannot open %s: %s\n", argv[i], strerror(errno));
} else {
printf("%s\n", argv[i]);
while ((c = getc(fp)) != EOF) {
putchar(c);
}
fclose(fp);
}
}
return 0;
}
int main(int argc, char *argv[]) {
FILE *fp;
int i;
char buff[128];
for (i = 1; i < argc; i++) {
printf("\n%s\n", argv[i]);
if(NULL == (fp = fopen(argv[i], "r"))){//check open file
perror("fopen");
continue;
}
while (fgets(buff, sizeof buff, fp)) {//read into buffer
fputs(buff, stdout);//print buffer (not add newline)
}
fclose(fp);
}
return 0;
}
I am quite new to C programming and I have to make a program that asks the user to enter the file name to open and then open that file and print the values sorted.
Will I need to pass the file pointer to main, or can I just open the file in one function and I can work with the file throughout the other functions?
int getFile ()
{
char file_name[100];
FILE* fp;
int rc;
printf("Enter the file name: ");
rc = scanf("%s", file_name);
if (rc != 1)
printf ("error");
fp = fopen(file_name, "r");
return 0;
}
Do I have to pass the file pointer from here to main?
Just pass the file name as an argument to main function, it maybe what you want.
You might use int main(int argc, char* argv[]), an example:
int main(int argc, char* argv[])
{
FILE* fp;
if (argc == 1)
printf("usage : a.out filename\n");
else
{
if (fp = fopen(*++argv, "r") != NULL)
{
/*your code here*/
}
}
return 0;
}
You can open the file in one function and pass it as argument to other functions. For example, for the pre-open file handle stdin, you could use it as:
char mystring [100];
fgets (mystring , 100 , pFile)