C Language Study Notes (1)

1. source code and supplemental code

Example

'6': source code: 0110; supplemental code: 0110

'-6': source code: 1110; reverse code: 1001; Complement: 1010 = 1001 + 1

(-6) + 6 = 0110 + 1010 = (1) 0000 = 0;

When the complement code is obtained, the original code is obtained;

2. compilation process:

Pre-compile --> compile (assembly language) --> assemble (machine language) --> Link

3. 'chmod-R': you can modify all permissions of subdirectories and files in the directory.

4. Comparison between snprintf () and sprintf (): snprintf () requires the second parameter to specify the size of the target buffer to ensure that the buffer does not overflow.

The general programming model of a 5.32-bit UNIX system is called the ILP32 model, which indicates that INTEGER (I), long integer (L), and pointer (P) Both occupy 32 bits.

Data Type ILP32 model LP64 Model

Char 8 8

Short 16 16

Int 32 32

Long 32 64

Pointer 32 64

6. the suffixes for type conversion of Integer constants include: u or U (unsigned), l or L (long), u/U and l/L combination (for example: ul, lu, Lu, etc ). For example, 100u;-123u; 0x123l.

Use the suffix L or l to forcibly store integers in the long type.

All real numbers are stored as double.

7. Storage of constants in memory:

1 (0x0001): stored in memory in unsigned short (two bytes;

'1' (0x31): one byte;

"1": The String constant must end with '\ 0', which occupies two bytes.

8. scanf () input end condition:

A. in case of space, TAB, or press ENTER

B.

C. Illegal characters

If scanf () has a return value, "1" is returned successfully. If it fails, "0" is returned ".

9.

A: for (I = 0; I <n; I ++)

A [I] = 4;

B: for (p = a; p <a + n; p ++)

* P = 4;

(A) Start With a and move the I elements forward. This is required for each assignment. "[]" is an operator.

(B) it is enough to move one unit behind each time. (B) The efficiency is significantly higher.

10.% f and % lf

　　

():

<Stdio. h>

Int main ()

{

Int a, B, c;

Double y;

A = scanf ("% lf", & y );

Printf ("% d, % f, % lf \ n", a, y, y );}

Enter 1234567891234567890

Output 1, 1234567891234567936.000000, 1234567891234567936.000000

Gdb:

Breakpoint 1, main () at 12_11_4.c: 7
7 a = scanf ("% lf", & y );
(Gdb) s
12345678912345678912
8 printf ("% d, % f, % lf \ n", a, y, y );
(Gdb) p y
$2 = 1.234567891234568e + 19

(B ):

# Include <stdio. h>

Int main ()

{

Int a, B, c;

Double y;

A = scanf ("% f", & y );

Printf ("% d, % f, % lf \ n", a, y, y );

}

Enter 123.11

Output 1, 0.000000, 0.000000

Why is the second output zero?

Check gdb debugging:

Breakpoint 1, main () at 12_11_4.c: 7
7 a = scanf ("% f", & y );
(Gdb) s
1234.123
8 printf ("% d, % f, % lf \ n", a, y, y );
(Gdb) p y
$1 = 4.8545332942524682e-270

1234.123 is read successfully, but the storage format is float.