为什么我的 C 程序没有打印出字符串？

Question

I am am making a program that prints out what the user types in, and it needs to be done with the method read_line() (for my homework), so I can't change too much.

I don't understand why it isn't printing out what the user enters.

#include <stdlib.h>

char *read_line(char *buf, size_t sz) {
  char tempBuf[sz];
  char c;
  int pos = 0;

  printf("> ");

  while(1) {
    c = getchar();
    if (tempBuf[pos] == EOF || tempBuf[pos] == '\n') {
      buf = tempBuf;
      return buf;
    } else {
      tempBuf[pos] = c;
    }
    pos++;
  }
}

int main(int argc, char **argv) {
  char *buf;
  char *input = read_line(buf, 128);

  printf("Here: %s", input);
}

I am new to C and I am finding it very confusing, so please explain anything in pretty simple terms. Any help would be much appreciated.

Answer 1

char *buf;
char *input = read_line(buf, 128);

Your first line creates a pointer variable called buf but does not assign it any particular value. Your second line passes the value of buf to read_line -- but you never assigned it any particular value. So you passed garbage to read_line and told it to use that garbage as a buffer.

You might want char buf[128]; instead of char *buf; , but it's hard to tell.

Also, please see my comment about your read_line function being broken in a way that indicates that whoever wrote it does not understand how to use getchar .

Answer 2

There are many mistakes in your program.

For starters you are passing to the function an uninitialized pointer

char *buf;
char *input = read_line(buf, 128);

In fact there is no sense to pass the pointer value of which is not used in the function.

Within the function the variable c should be declared as having the type int.

int c;

Otherwise if the type char behaves as the type unsigned char (it depends on a compiler option) a comparison it with EOF will always evaluates to false.

Within the function the array tempBuf is not initialized. So this if statement

if (tempBuf[pos] == EOF || tempBuf[pos] == '\n') {

invokes undefined behavior.

Within the wjile llop you have to check that the value of pos is less than the value of sz .

The function returns a pointer to a local array that makes the returned pointer invalid.

  buf = tempBuf;
  return buf;

Moreover the array even does not contain a string because the terminating zero is not appended to the array.

The function should allocate dynamically memory and return pointer to the allocated memory that will contain a zero terminated string.

Below there is a demonstrative program that shows how the function can be written.

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

char * read_line( size_t n ) 
{
    char *s = malloc( n );

    if ( s != NULL )
    {
        int c;

        printf( "> " );

        size_t i = 0;

        for ( ; i + 1 < n && ( c = getchar() ) != EOF && c != '\n'; i++ )
        {
            s[i] = c;
        }

        s[i] = '\0';
    }

    return s;
}

int main(void) 
{
    size_t n = 128;

    char *input = read_line( n );

    if ( input != NULL ) printf( "Here: %s\n", input );

    free( input );

    return 0;
}

the program output might look like

> Hello Jake Jackson
Here: Hello Jake Jackson

Answer 3

The current implementation of read_line is very flawed due to the simple fact that it returns a locally declared buffer (which gets deleted at the end of the function). As a result, your pointers are pointing at garbage values (which is very dangerous - I repeat - as it will cause your program to crash or worse [keep running even when using memory it doesn't own]).

What you should be doing instead a dynamically creating the buffer on the heap (that way it can safely be returned - but it would have to be manually deleted).

So your function (and the rest of the code) should change to something like the following:

#include <stdlib.h>

char *read_line(size_t sz) {
  // create a new buffer on the heap - so it can easily be returned (note this will have to be deleted after use using free()).
  char *tempBuf = malloc(sz);
  int c;
  int pos = 0;

  // set all elements of tempBuf to nulls
  memset(arr, 0, sz); 

  printf("> ");

  while(1) {
    c = getchar();
    if (tempBuf[pos] == EOF || tempBuf[pos] == '\n') {
      return tempBuf;
    } else {
      tempBuf[pos] = (char) c;
    }
    pos++;
  }
}

int main(int argc, char **argv) {
  char *input = read_line(128);    
  printf("Here: %s", input);

  // free the memory since we are now done with it.
  free(input);
}

You can read more about the heap & stack here .

Answer 4

There are a few issues with your program.

In your main() , you initialize buf , but never allocate any memory for it. You can allocate memory using malloc :

    char *buf = malloc(128);
    char *input = read_line(buf, 128);

In read_line() , you initialise tempBuf , but read an element from it before initialising the element. tempBuf holds junk values, which you use for comparisons. You initialise the element after the comparison, which is not correct. There are other issues as well.

Here is a modified version with minimal changes to your code:

char *read_line(char *buf, size_t sz) {
  char *tempBuf = malloc(sz);
  char c;
  int pos = 0;

  printf("> ");

  while(1) {
    c = getchar();
    if (c == '\n') {
      tempBuf[pos] = '\0';
      buf = tempBuf;
      return buf;
    } else {
      tempBuf[pos] = c;
    }
    pos++;
  }
}

The last character of any string in C needs to be a null character ('\0').

Tip: You should also add a check for pos , so it doesn't exceed the value of sz .

Answer 5

Try this one:

#include<stdio.h>       //originally missing - needed for I/O
#include<string.h>      //for memset

void read_line(char *buf, size_t sz) 
{
   char c;
   int pos = 0;

   printf("> ");

   while(pos<sz-1) 
   {
       c = getchar();
       if (c == EOF || c == '\n')  //originally tempBuf[pos] was used which contained garbage value
       {
           break;
       } 
       else 
       {
           buf[pos] = c;
       }
       pos++;
   }
}

 int main(int argc, char **argv) 
 {
    char buf[128];             //buf is now an array instead of char*
    memset(buf,'\0',128);
    read_line(buf, 128);    //buf is the array in this code as the array tempBuf is local and will be destroyed once we come out of read_line function 

    printf("Here: %s", buf);
 }

Errors have been commented in this code.

Answer 6

You have so many errors. Local variables used outside the scope, not terminated strings etc etc. Here you have working example https://godbolt.org/z/_3ZHUJ

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

char *read_line(size_t sz) 
{
  char *tempBuf = malloc(sz);
  char c;
  int pos = 0;

  printf("> ");

  while(1) {
    c = getchar();
      tempBuf[pos] = c;
    if (tempBuf[pos] == EOF || tempBuf[pos] == '\n') 
    {
      tempBuf[pos] = 0;
      return tempBuf;
    }
    pos++;
  }
}

int main(int argc, char **argv) {
  char *buf;
  char *input = read_line( 128);

  printf("Here: %s", input);
}