second print statement goes wrong

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

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.

Quote:

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

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.

Quote:

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.

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; }

Quote:

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

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.

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.