second print statement goes wrong

**Dadu@** · 03-06-2022

I am looking help to perform following operation

Data_Byte = 0000 0000 | 0000 0001 = 0000 0001 << 1 = 000 00010 = 2

Data_Byte = 000 00010 | 0000 0001 = 0000 0011 << 1 = 000 00110 = 6

User can modify LSB bit

My last print statement should print value 6 as expected. I don't understand why I don't get 6. What I need to print 6 ?

Code:

#include<stdio.h>

#include<stdint.h>

int main ()
{
    uint8_t Data_Byte = 0;
    uint8_t LSB_BIT;
    
    printf("Set/Reset LSB Enter 1 /0 : ");
    scanf("%d", &LSB_BIT );
 
    Data_Byte =  ( Data_Byte | LSB_BIT );
    printf(" %d\n",  Data_Byte );
 
    Data_Byte = Data_Byte << 1;
    printf("Byte =  %d \n",  Data_Byte );
    
    printf("Set/Reset LSB Enter 1 /0 : ");
    scanf("%d", &LSB_BIT );
 
    Data_Byte =  ( Data_Byte | LSB_BIT );
    printf("%d\n",  Data_Byte );
 
    Data_Byte = Data_Byte << 1;
    printf("Byte =  %d \n",  Data_Byte );
 
   return 0;
}

Set/Reset LSB Enter 1 /0 : 1
1
Byte = 2
Set/Reset LSB Enter 1 /0 : 1
1
Byte = 2

**Salem** · 03-07-2022

Well here's the first problem.

Code:

foo.c: In function ‘main’:
foo.c:11:13: warning: format ‘%d’ expects argument of type ‘int *’, but argument 2 has type ‘uint8_t *’ {aka ‘unsigned char *’} [-Wformat=]
   11 |     scanf("%d", &LSB_BIT );
      |            ~^   ~~~~~~~~
      |             |   |
      |             |   uint8_t * {aka unsigned char *}
      |             int *
      |            %hhd
foo.c:20:13: warning: format ‘%d’ expects argument of type ‘int *’, but argument 2 has type ‘uint8_t *’ {aka ‘unsigned char *’} [-Wformat=]
   20 |     scanf("%d", &LSB_BIT );
      |            ~^   ~~~~~~~~
      |             |   |
      |             |   uint8_t * {aka unsigned char *}
      |             int *
      |            %hhd

Your scanf with %d is going to trash memory outside of the intended variable.

Using the correct scanf format of %hhd, I get

Code:

$ gcc foo.c
$ ./a.out 
Set/Reset LSB Enter 1 /0 : 1
 1
Byte =  2 
Set/Reset LSB Enter 1 /0 : 1
3
Byte =  6

Enable as many warnings as you can on your compiler, and use a compiler with lots of diagnostics.

**Dadu@** · 03-07-2022

Originally Posted by Salem

Well here's the first problem.

Using the correct scanf format of %hhd, I get
.

I don't get correct value even after using %hhd,

Code:

 #include<stdio.h>

#include<stdint.h>


int main ()
{
    uint8_t Data_Byte = 0;
    uint8_t LSB_BIT;
    
    printf("Set/Reset LSB Enter 1 /0 : ");
    scanf("%hhd", &LSB_BIT );
 
    Data_Byte =  ( Data_Byte | LSB_BIT );
    printf(" %d\n",  Data_Byte );
 
    Data_Byte = Data_Byte << 1;
    printf("Byte =  %d \n",  Data_Byte );
    
    printf("Set/Reset LSB Enter 1 /0 : ");
    scanf("%hhd", &LSB_BIT );
 
    Data_Byte =  ( Data_Byte | LSB_BIT );
    printf("%d\n",  Data_Byte );
 
    Data_Byte = Data_Byte << 1;
    printf("Byte =  %d \n",  Data_Byte );
 
   return 0;
}

warnings

hello.c: In function 'main':
hello.c:11:14: warning: unknown conversion type character 'h' in format [-Wformat=]
scanf("%hhd", &LSB_BIT );
^
hello.c:11:11: warning: too many arguments for format [-Wformat-extra-args]
scanf("%hhd", &LSB_BIT );
^~~~~~
hello.c:20:14: warning: unknown conversion type character 'h' in format [-Wformat=]
scanf("%hhd", &LSB_BIT );
^
hello.c:20:11: warning: too many arguments for format [-Wformat-extra-args]
scanf("%hhd", &LSB_BIT );

output

Set/Reset LSB Enter 1 /0 : 1
1
Byte = 2
Set/Reset LSB Enter 1 /0 : 1
1
Byte = 2

**Salem** · 03-07-2022

I thought you were using GCC.

Which version?

I have

Code:

$ gcc --version
gcc (Ubuntu 9.3.0-17ubuntu1~20.04) 9.3.0
Copyright (C) 2019 Free Software Foundation, Inc.
This is free software; see the source for copying conditions.  There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.

Maybe start with
int Data_Byte = 0;
int LSB_BIT;

and regular vanilla %d to make sure it's OK.

**Dadu@** · 03-07-2022

Originally Posted by Salem

I thought you were using GCC.

Which version?

I have
[code]
$ gcc --version
gcc (Ubuntu 9.3.0-17ubuntu1~20.04) 9.3.0
Copyright (C) 2019 Free Software Foundation, Inc.
This is free software; see the source for copying conditions. There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
.

I have

Code:

 gcc --versiongcc (MinGW.org GCC-6.3.0-1) 6.3.0
Copyright (C) 2016 Free Software Foundation, Inc.
This is free software; see the source for copying conditions.  There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.

**flp1969** · 03-07-2022

inttypes.h header gives us the proper format conversion strings:

Code:

#include <stdio.h>
#include <stdint.h>
#include <inttypes.h>

int main ()
{
  uint8_t Data_Byte = 0;
  uint8_t LSB_BIT;

  printf ( "Set/Reset LSB Enter 1 /0 : " );
  scanf ( "%" SCNu8, &LSB_BIT );

  Data_Byte =  ( Data_Byte | LSB_BIT );
  printf ( "%" PRIu8 "\n",  Data_Byte );

  Data_Byte = Data_Byte << 1;
  printf ( "Byte =  %" PRIu8 "\n",  Data_Byte );

  printf ( "Set/Reset LSB Enter 1 /0 : " );
  scanf ( "%" SCNu8, &LSB_BIT );

  Data_Byte =  ( Data_Byte | LSB_BIT );
  printf ( "%" PRIu8 "\n",  Data_Byte );

  Data_Byte = Data_Byte << 1;
  printf ( "Byte =  %" PRIu8 "\n",  Data_Byte );

  return 0;
}

**Dadu@** · 03-07-2022

Originally Posted by flp1969

inttypes.h header gives us the proper format conversion strings:

I ran your code but I get same result as previous

Code:

  hello.c: In function 'main':hello.c:11:11: warning: unknown conversion type character 'h' in format [-Wformat=]
   scanf ( "%" SCNu8, &LSB_BIT );
           ^~~
hello.c:11:11: warning: too many arguments for format [-Wformat-extra-args]
hello.c:20:11: warning: unknown conversion type character 'h' in format [-Wformat=]
   scanf ( "%" SCNu8, &LSB_BIT );
           ^~~
hello.c:20:11: warning: too many arguments for format [-Wformat-extra-args]

Set/Reset LSB Enter 1 /0 : 1
1
Byte = 2
Set/Reset LSB Enter 1 /0 : 1
1
Byte = 2

**laserlight** · 03-07-2022

Since you're using MinGW, the issue here might be that the Visual C runtime is used, or something like that. It is why instead of the standard %lld you may have to use %I64d or something like that to print long long, and something like that may be coming into play here.

**christop** · 03-08-2022

I don't see any strong reason to use uint8_t for LSB_BIT. You could change it to unsigned and then use %u for the scanf format. You wouldn't even have to make any other changes in your code.

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