To read the content of a file in C - c

Question

I have a C code to read a txt file:
#include <stdio.h>
int main()
{
FILE *pf;
int ii;
int jj;
char *filename;
printf("enter file name");
scanf("%s",filename);
printf("%s",filename);
pf = fopen("filename+.txt", "r");
if(pf==Null)
{
printf("cant open");
}
else
{
fscanf(pf,"%d,%d" ,&ii,&jj );
printf("%d,%d\n" ,ii,jj);
}
fclose(pf);
return 0;
}
Still i get segmentation error.
The input txt file contains
2,3
I get segmentation fault(core dumped) when i run the program as ./readfile input.
What is going wrong here , how can i correct this?

Answer 1

int main(char *) is not a legal signature for main in C. Only
int main(void)
and
int main(int argc, char **argv)
are legal. In your case, you will need the latter.

Answer 2

That is not the correct way to specify arguments to your program. ie you can't do this:
int main(char *filename)
There should have been a compiler error when you compiled your program. The correct definition is:
int main( int argc, char **argv )
Where argv is an array of strings. Try doing this experiment:
int main( int argc, char **argv )
{
int i;
for( i = 0; i < argc; i++ ) {
printf( "arg %d is: \"%s\"\n", argv[i] );
}
return 0;
}
Then, write your program to use the correct argument list as above.
One other point to make is that you should test the return value of fopen. If it is NULL, then you should NOT try to access the file (because it failed to open).

Answer 3

There are many mistakes here.
After calling fopen(), you should check if pf is NULL, because fopen() can fail.
You are trying to open the file of name filename+.txt. Shouldn't you be opening the file which the name was provided as parameter?
Also, the structure of main() should be int main(int argc, char **argv), you cannot do whatever you want about this.
Check if argc > 1, in which case the program was started with parameters, and the file name should've been provided in argv[1].
Update on comments: This is how your code should look like:
int main()
{
char filename[512]; // reserve 512 bytes to receive the file name from input
FILE *pf;
int ii;
int jj;
printf("Enter file name: ");
scanf("%s", filename);
pf = fopen(filename, "r");
if (pf)
{
fscanf(pf,"%d,%d", &ii, &jj);
printf("%d,%d\n", ii, jj);
fclose(pf);
}
else
{
printf("Failed to open file name %s", filename);
}
return 0;
}
You can also do this to get the filename from the parameters:
int main(int argc, char **argv)
{
FILE *pf;
int ii;
int jj;
if (argc > 1)
{
pf = fopen(argv[1], "r");
if (pf)
{
fscanf(pf, "%d,%d", &ii, &jj);
printf("%d,%d\n", ii, jj);
fclose(pf);
}
else
{
printf("Failed to open file name %s", argv[1]);
}
}
else
{
printf("Insuficient parameters");
}
return 0;
}
Or even, if you don't want to pass the file extension:
int main(int argc, char **argv)
{
char *filename;
FILE *pf;
int ii;
int jj;
if (argc > 1)
{
filename = malloc(strlen(argv[1]) + 5); // alloc necessary memory
strcpy(filename, argv[1]);
strcat(filename, ".txt");
pf = fopen(filename, "r");
if (pf)
{
fscanf(pf, "%d,%d", &ii, &jj);
printf("%d,%d\n", ii, jj);
fclose(pf);
}
else
{
printf("Failed to open file name %s", filename);
}
}
else
{
printf("Insuficient parameters");
}
return 0;
}