Printing part of a string

This is a discussion on Printing part of a string within the C Programming forums, part of the General Programming Boards category; Hi, I've a string str as: HiHello I want to print only the Hello part i.e. from index 2 till ...

  1. #1
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    Printing part of a string

    Hi,

    I've a string str as:

    HiHello

    I want to print only the Hello part i.e. from index 2 till the end like
    print(str[3...endofstr]

    How can I do that?

    Thanks,
    Angkar

  2. #2
    Awesomefaceradcore bivhitscar's Avatar
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    Code:
    for ( i = 2; i <= strlen( str ); i++ )
    {
    printf("%c", str[i] );
    }
    That should do the trick.

    p.s. sorry about the lack of tabs in my code

  3. #3
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    And what if I want to store the string str[3...endofstr] in some other variable?

  4. #4
    Registered User 00Sven's Avatar
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    Code:
    for( i=0, d=3; d < strlen( str ); i++,d++){
         str2[i]=str1[d];
    }
    ~Sven
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    On an output stream, fflush causes any buffered but unwritten data to be written; On an input stream, the effect is undefined. It returns EOF for a write error, and zero otherwise. fflush(NULL) flushes all output streams.
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  5. #5
    Registered User Mortissus's Avatar
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    Or you could make:
    Code:
    {
       strncpy(str2,&str1[2],str2_size);
    }

  6. #6
    int x = *((int *) NULL); Cactus_Hugger's Avatar
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    Quote Originally Posted by bivhitscar
    Code:
    for ( i = 2; i <= strlen( str ); i++ )
    {
    printf("%c", str[i] );
    }
    That should do the trick.

    p.s. sorry about the lack of tabs in my code
    You can lose the for loop on that, if you're going to use printf():
    Code:
    printf("%s", str + 2);
    Of course, you should do bounds checking.
    Code:
    if(strlen(str) >= 2) printf("%s", str + 2);
    If you want to store it, another solution is:
    Code:
    sprintf(store_str, "%s", str + 2);
    Edit: Note that the "%s" isn't needed in any of those, but if you leave it out, someone can type something like "blah %d %d", and trick s/printf().
    Last edited by Cactus_Hugger; 04-23-2006 at 07:47 PM.
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  7. #7
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    Or you could do something like this.

    Code:
    #include <stdio.h>
    #include <string.h>
    #include <stdlib.h>
    
    int main(int argc, char **argv) {
        if(argc !=2)
            exit(1);
    
        char temp[]="Hello";
    
        size_t len_temp = strlen(temp);
        size_t len_argv = strlen(argv[1]);
        size_t len_diff = len_argv - len_temp;
    
        if(len_diff != 2)
            exit(1);
    
        if( strstr(argv[1],temp) )
            printf("%s\n",temp);
    
        return 0;
    }
    $gcc -Wall match.c -o match
    $./match HiHello
    Hello
    $./match MeHello
    Hello
    $./match NoooHello
    $./match Hello
    $./match ribbit
    $./match Hell
    $

    Of course, the code is case sensitive.

  8. #8
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    Thanks to CDALTEN as his example taught me new thing i.e. size_t . It's really cool to use it instead of making new charecter type arrays.



    REALNAPSTER

  9. #9
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    Quote Originally Posted by 00Sven
    Code:
    for( i=0, d=3; d < strlen( str ); i++,d++){
         str2[i]=str1[d];
    }
    ~Sven
    This will call strlen(str) for every iteration and so is much slower than if you only call strlen once.
    Code:
    int len = strlen(str);
    
    for( i=0, d=3; d < len; i++,d++){
         str2[i]=str1[d];
    }

  10. #10
    ATH0 quzah's Avatar
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    I think you're not understanding what they're doing. He's simply using a variable of type size_t to store the return value from the functions. Said functions return that type. It has nothing to do with "making new character type arrays" or not.


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

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