Cut first letter of output with fgets in C

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Hello.
I want to give an input via fgets() ,but I'm in trouble.

For example, I give :

hello and I receive ello
I give hello world I receive ello world
How to solve it?

Thanks!!!

Code:

What I have tried:

C++

<pre>#include <stdio.h>
#include <stdlib.h>


char *input();

int main(void)
{
    while(1)
    {
        char *a = input();
        printf("%s..." , a);
    }
}

char *input()
{
    printf("Give: ");
    char *word = malloc(sizeof(char) * 50);
    fgetc(stdin);   
    fgets(word , 100 , stdin); 
    return word;
}

I have change the code a little bit (in my basic program (that's was just a test)) ,and now is like this:

C++

char *word = malloc(sizeof(char) * 100);

fgets(word , sizeof(word) , stdin);   /
while (fgets(word,sizeof(word),stdin) != NULL)
{
    break;
}
size_t len = strlen(word) - 1;
if (word[len] == '\n') //without newline
{
    word[len] = '\0';
}
return word;

It works perfect for simple words ,but when I give a sentence ,for example :
hello world ,I receive hello w (without "orld").......

Posted 26-Sep-20 4:47am

Nick_is_asking

Updated 26-Sep-20 8:10am

v4

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3 solutions

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Patrice T · Answer 1 · 2020-09-26T04:59:00

Quote:
hello and I receive ello

C++

char *input()
{
    printf("Give: ");
    char *word = malloc(sizeof(char) * 50);
    fgetc(stdin);   // may be this is what eating the first letter
    fgets(word , 100 , stdin); 
    return word;
}

[Update]
Try to replace:

C++

fgets(word , sizeof(word) , stdin);

with:

C++

fgets(word , 100 , stdin);

word is a pointer, its size is the size of the pointer, not the size of space it is pointing to.
In C/C++, arrays have np knowledge of their own size.

Rick York · Answer 2 · 2020-09-26T08:10:00

Solution 3

Here's how I would write that code :

C++

void input( char buffer, size_t bufferSize )
{
    char * p;
    printf("Give: ");
    fgets( buffer, bufferSize, stdin );
    p = strchr( buffer, '\n' );
    if( p )
       *p = 0;     // null new line character if found
}

#define BUFFER_SIZE  79

int main(void)
{
    char buffer[ BUFFER_SIZE + 1 ] = { 0 };   // +1 for terminating null
    while(1)
    {
        input( buffer, BUFFER_SIZE );         // leave room for the null
        printf( "input was '%s'\n" , buffer );
        if( stricmp( buffer, "exit" ) == 0 )
            break;                            // give a way to escape the loop
    }
    return 0;
}

Posted 26-Sep-20 8:10am

Rick York

Comments

Nick_is_asking 26-Sep-20 15:07pm

I test it and it works, BUT when I try to do it in my basic code I receive: Segmentation fault (core dumped)

char *get_word(char *buffer , size_t bufferSize , char *player_A , char *player_B , int flag)
{

char *p;

//fgetc(stdin);
fgets( buffer, bufferSize, stdin );
p = strchr( buffer, '\n' );
if( p )
{
*p = 0;
}

return buffer;
}

In main I have this: char word[ 100 + 1 ] = { 0 }; and I call
char *str = get_word(word , 100 , A , B , 0);
but doesn't work...

Rick York 26-Sep-20 15:39pm

You are passing A and B in but nothing is done with them so why are you passing them?
Why bother returning the address of the buffer passed in when it is already known?

Nick_is_asking 26-Sep-20 15:49pm

Because I want to return this character to others functions(Program is big).
I tried MANY methods ,but still nothing...

Nick_is_asking · Answer 3 · 2020-09-26T05:20:00

Solution 2

No,it doesn't work.
I have change the code a little bit (in my basic program (that's was just a test)) ,and now is like this:

C++

char *word = malloc(sizeof(char) * 100);

fgets(word , sizeof(word) , stdin);   /
while (fgets(word,sizeof(word),stdin) != NULL)
{
    break;
}
size_t len = strlen(word) - 1;
if (word[len] == '\n') //without newline
{
    word[len] = '\0';
}
return word;

It works perfect for simple words ,but when I give a sentence ,for example :
hello world ,I receive hello w (without "orld").......

Posted 26-Sep-20 5:20am

Nick_is_asking

Comments

Richard MacCutchan 26-Sep-20 11:29am

fgets(word , sizeof(word) , stdin);
The sizeof operator is a compile time operation, and gives the size of the object referred to. In your case sizeof(word) will give the answer 4, since word is a pointer. So you are telling fgets that your buffer is only 4 bytes wide.

[edit]
I guess you are building 64-bit as the size of your pointer would appear to be 8 bytes.
[/edit]

Nick_is_asking 26-Sep-20 11:32am

Thanks for the update.
so ,what can I do?

Richard MacCutchan 26-Sep-20 11:38am

Get a book on the C language.

But the quick fix to your code is:

#define BUFLEN 100
char *word = (char*)malloc(sizeof(char) * BUFLEN);

fgets(word, BUFLEN, stdin);
while (fgets(word, BUFLEN, stdin) != NULL)
{
// what is this loop for?
}

Nick_is_asking 26-Sep-20 11:55am

I'm a college student.I will have C this semester,so they will give us book.
As about this loop ,I found it on internet.

Richard MacCutchan 26-Sep-20 12:03pm

Stop wasting your time copying code from the internet which you do not understand. It will only serve to confuse you when you try to learn the language properly.