Using fgets, removing last character

This is a discussion on Using fgets, removing last character within the C Programming forums, part of the General Programming Boards category; Hello, I have ran into a slight problem. Of course I always did it the bad way I guess and ...

  1. #1
    C Programmer Stack Overflow's Avatar
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    Using fgets, removing last character

    Hello,

    I have ran into a slight problem. Of course I always did it the bad way I guess and now I'm looking for a solution. Here is my current code:

    Code:
    int main() {
    	char sentence[256];
    
    	printf("Enter your sentence here: ");
    	fgets(sentence, 255, stdin);
    	sentence[strlen(sentence)-1] = '\0';
    	
    	printf("Sentence: %s\n", sentence);
    	
    	return 0;
    }
    The only thing is, I don't like calling strlen() in that situtation. I've tried everything from *(sentence--) = '\0' to sentence[*sentence-1] = '\0' and I just can't seem to find a way.

    To keep an open mind, is there actually a way or is this [current way] the best way?


    Thank you for your time,
    - Stack Overflow
    Segmentation Fault: I am an error in which a running program attempts to access memory not allocated to it and core dumps with a segmentation violation error. This is often caused by improper usage of pointers, attempts to access a non-existent or read-only physical memory address, re-use of memory if freed within the same scope, de-referencing a null pointer, or (in C) inadvertently using a non-pointer variable as a pointer.

  2. #2
    Crazy Fool Perspective's Avatar
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    fgets will automatically append the '\0' so long as there is no error in reading.

  3. #3
    ATH0 quzah's Avatar
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    Quote Originally Posted by Perspective
    fgets will automatically append the '\0' so long as there is no error in reading.
    To add to this, this is why you should check the return value of your functions...

    Quzah.
    Hope is the first step on the road to disappointment.

  4. #4
    C Programmer Stack Overflow's Avatar
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    I agree,

    Though in fact fgets returns a char * and when dealing with the function, or maybe even in Linux itself, my sentence I retrieve retrieves the '\n' at the end of the string for some strange reason.

    I will look into this with more detail.

    - Stack Overflow
    Segmentation Fault: I am an error in which a running program attempts to access memory not allocated to it and core dumps with a segmentation violation error. This is often caused by improper usage of pointers, attempts to access a non-existent or read-only physical memory address, re-use of memory if freed within the same scope, de-referencing a null pointer, or (in C) inadvertently using a non-pointer variable as a pointer.

  5. #5
    & the hat of GPL slaying Thantos's Avatar
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    Well a couple things to remember
    1) fgets will get the '\n' if there is room
    2) fgets will always append a null character
    3) strlen returns the number of characters not counting the null character.

    So if the input what "able" your string would be:
    "able\n\0", strlen would return 5, 5 - 1 = 4, string[4] is '\n'

  6. #6
    C Programmer Stack Overflow's Avatar
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    I see,

    Thanks. A program I'm writing doesn't like the whole '\n' character, and my algorithm usually gets messed up if that happens, so I'm going to take another approach:

    Code:
    int main() {
    	char sentence[256], *sent;
    
    	printf("Enter your sentence here: ");
    	sent = fgets(sentence, 255, stdin);
    	sent[strlen(sentence) - 1] = '\0';
    	
    	printf("Sentence: %s\n", sent); // use sent instead of sentence from now on
    	
    	return 0;
    }
    That shouldn't conflict with sentence now, and the whole length of the string recursively calling itself. I thought there would be an easier way, but I guess this is the best way around it for now.


    Thank you for your time,
    - Stack Overflow
    Segmentation Fault: I am an error in which a running program attempts to access memory not allocated to it and core dumps with a segmentation violation error. This is often caused by improper usage of pointers, attempts to access a non-existent or read-only physical memory address, re-use of memory if freed within the same scope, de-referencing a null pointer, or (in C) inadvertently using a non-pointer variable as a pointer.

  7. #7
    Registered User linuxdude's Avatar
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    the reason that
    Code:
    *(sentence--) = '\0
    doesn't work is because the memory address is at the beggining for strings. So sentence-- will be one before the actual string. It really is the best way, because either way you will have to get to the end of a string. if you want you could use (with caution)scanf
    Code:
    #include <stdio.h>
    #include <stdlib.h>
    
    int main(void){
            char string[100];
            printf("Enter a string\n");
            if(!scanf("%100s*[^\n]",string)){
                    printf("Failure\n");
                    return EXIT_FAILURE;
            }
            printf("%s",string);
            return EXIT_SUCCESS;
    }
    this won't get spaces though
    Last edited by linuxdude; 07-05-2004 at 08:55 PM.

  8. #8
    ATH0 quzah's Avatar
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    Code:
    if( (c=strrchr( buf, '\n' )) )
        *c='\0';
    With, or without the r.

    Quzah.
    Hope is the first step on the road to disappointment.

  9. #9
    Registered User
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    forgive me my ignorance but, in this case wouldnt it be easier to use gets instead of fgets?

    Even though it dosnt have built in array overrun you could always just code in an if statement to tell it the last chr is null if it exceeds size and then you wouldnt have to worry about \n being in there and save your self a few clock cycles.

    ~Meloshski

  10. #10
    Obsessed with C chrismiceli's Avatar
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    To quote the header file I got with gcc
    Quote Originally Posted by stdio.h
    /* Get a newline-terminated string from stdin, removing the newline.
    DO NOT USE THIS FUNCTION!! There is no limit on how much it will read.
    That is in reference to gets, always use fgets over gets.
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  11. #11
    & the hat of GPL slaying Thantos's Avatar
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    gets() is the one thing I have no problem saying: NEVER EVER USE IT. THERE IS NO EXCUSE FOR USING gets().

  12. #12
    ATH0 quzah's Avatar
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    Even though it dosnt have built in array overrun you could always just code in an if statement to tell it the last chr is null if it exceeds size and then you wouldnt have to worry about \n being there
    Yeah, because you'd be too busy worrying about it crashing your program or causing a nice little security problem because someone used a buffer overrun to trash you OS or ...

    Quzah.
    Hope is the first step on the road to disappointment.

  13. #13
    C Programmer Stack Overflow's Avatar
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    Thanks everyone,

    I too agree there is no possible excuse to use gets() for any compilation. Why they created these functions, hence gets() and fflush(), still comes unclear to me.

    I like the method:
    Code:
    if( (c=strrchr( buf, '\n' )) )
        *c='\0';
    It's the fastest, safest, and most reasonable approach in my view to remove the '\n' character from the string.


    Once again thank you all, and thank you quzah,
    - Stack Overflow
    Segmentation Fault: I am an error in which a running program attempts to access memory not allocated to it and core dumps with a segmentation violation error. This is often caused by improper usage of pointers, attempts to access a non-existent or read-only physical memory address, re-use of memory if freed within the same scope, de-referencing a null pointer, or (in C) inadvertently using a non-pointer variable as a pointer.

  14. #14
    ATH0 quzah's Avatar
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    fflush is a useful function, however, people always use it wrong around these parts, trying to flus input streams instead of output...

    Quzah.
    Hope is the first step on the road to disappointment.

  15. #15
    & the hat of GPL slaying Thantos's Avatar
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    Well fflush() has its usages like flushing the output buffer. Most likely gets() was implamented before fgets(). Now the question is why the still allow it to live

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