Thread: C Functions and varibles

im working on a little text to html conversion program. i have a function to create a title(1 param), function to chacge inputted name into a output-able name(t param, old filename and extension) and the function that does the work( 3 params, input file, output file and the title). when ever i call the main function using separte functions, all the varibles are the same. why!?!?!?!?
i have tried this a couple of different ways, and the funny part is it has worked in a function before(a function to chage the inputted filename to a output-able file).

heres some unfinished code of it
==============================================
# include <stdio.h>
# include <string.h>
# include <ctype.h>

int main(int argc, char *argv[]) {
process(argv[1], convertfileext(argv[1], ".html") );
}

int convertfileext(char name[], char ext[]) {
int x = strlen(name);
int converted = 0;

while(name[x] != EOF) {
if(name[x] == '.') {
name[x] = '\0';
name = strcat(name, ext);

converted = 1;
break;
}

--x;
}

if(converted == 0)
name = strcat(name, ext);

return name;
}

int createtitle(char filename[]) {
int x = 0;

filename[0] = toupper(filename[0]);
while(filename[x] != EOF) {
if(filename[x] == '_')
filename[x] = ' ';
else if(filename[x] == ' ')
filename[x+1] = toupper(filename[x+1]);

++x;
}

return filename;
}

int process(char path[], char newfile[]) {
printf("\n%s\n", path);
printf("%s\n", newfile);

return 0;
}

==============================================
the process function prints newfile out twice for me. i tried this on windows xp and in red hat linux 7.3, what is happening here!?!?!?

thanks

What if you had passed argv[0] to process() and convertfileext()?

Originally posted by The Dog
What if you had passed argv[0] to process() and convertfileext()?

argv[0] is the executable that's running itself.

thanx. i never use command line args.

so does any one know why all my varibles end up the same? i have read some books on C before and it been done there, it even done in one of mu functions. what is going on?

> what is going on?
You're modifying argv[1], and printing it out - there is only one string present here, so it's going to always print the same answer.

Also, modifying argv[i] strings is not something I would normally recommend.

Code:

# include <stdio.h> 
# include <string.h> 
# include <ctype.h> 

// function prototypes are your friends, get to know them
char *convertfileext(char name[], char ext[]);
void process(char path[], char newfile[]);

int main(int argc, char *argv[]) {
    char    new[100];
    strcpy( new, argv[1] );
    process(argv[1], convertfileext(new, ".html") ); 
} 

char *convertfileext(char name[], char ext[]) {
    char    *p;
    p = strrchr( name, '.' );   // find the last .
    if ( p != NULL ) {
        strcpy( p, ext );       // replace current .ext with the new one
    } else {
        strcat( name, ext );    // no .ext, so just add the new one
    }
    return name; 
} 

void process(char path[], char newfile[]) { 
    printf("\n%s\n", path); 
    printf("%s\n", newfile); 
}

i thought when you pass a arg to a function only a copy goes to the function, why isnt using argv[1] a good idea.

thanks for the post

> i thought when you pass a arg to a function only a copy goes to the function
It is when the thing you are passing to a function isn't an array. Strings are arrays of char, so all you get is a pointer to the start of the array.

> why isnt using argv[1] a good idea.
Because you've already fallen into the trap - you modified it, and got confused.

Also, you don't know how the memory is allocated. Making a string shorter is OK, but making it longer (with say strcat) is a definite NO, since some of the chars will end up being written to someplace you don't own.

thanks alot for the info. its been bugging me for awhile

sorry to bug you guys again but any thing in my program just doesnt want to work. i want to take input in allmost like a shell to a function but when a special word is typed in the program exits. heres what i have(yes, all the function work fine but my error message is still showed(function error()).
================================================
int main(int argc, char *argv[]) {
char file[255];
char file2[255];

if(argv[1] != (NULL)) {
int x = 1;

while(argv[x] != NULL) {
strcpy(file, argv[x]);
strcpy(file2, argv[x]);

printf("Converting %s ", argv[x]);
if( (process(argv[x], convertfileext(file, ".html"), createtitle(file2))) == 0)
printf("...Done.\n");

++x;
}
} else {
char input[255];
char exitnow[5] = "exit";

error();

while((strcmp(input, exitnow)) != 0) {
scanf("%s", input);
strcpy(file, input);
strcpy(file2, input);

printf("Converting %s ", input);
if( (process(input, convertfileext(file, ".html"), createtitle(file2))) == 0)
printf("...Done.\n");
}


}
}
================================================
shouldnt the code that could call the function never used if exit it typed?

i forgot to mention that function process calls error() if the file isnt found/readable. but still why is error being called?

thanks

i just noticed, it still doesnt work. i just changed the upper code to what you showed me. that one error call should be there so the user is told what to do. why does the while loop not stop when strcmp doesnt return 0?

by the way why wouldnt you use scanf? i new, just wondering.

thanks again guys for your posts