Casting in C++

Good question. Let’s break it clearly:

int*

• Declares a pointer to int.

• Example:

  int* p;  // p can point to an integer

(int*)

• This is a type cast operator.

• It tells the compiler: “Treat this value as a pointer to int”.

• Example with malloc:

  int* p = (int*) malloc(sizeof(int));

Here, malloc returns a void* (generic pointer). In C:

• void* can be assigned to any pointer type without cast.

• So this works fine too:

  int* p = malloc(sizeof(int));

In C++:

• Implicit conversion from void* to other pointers is not allowed.
• That’s why we must use (int*)malloc(...) in C++.

Summary

• In C: Cast is optional → int* p = malloc(sizeof(int)); (preferred).
• In C++: Cast is required → int* p = (int*)malloc(sizeof(int));
• Modern C++ avoids malloc and uses new/delete.

Do you want me to also show the difference between malloc+free vs new+delete in C++?

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safe conversions between integers and characters in C++,

1. Core Idea

• char is just a 1-byte integer
• Characters like '0', '1' are stored using ASCII values

Character	ASCII Value
'0'	48
'1'	49

2. Safe Conversions

Integer → Character (digit)

char c = digit + '0';

✔ Works when digit is between 0–9

Example:

int x = 1;
char c = x + '0'; // '1'

Character → Integer

int digit = c - '0';

Example:

char c = '1';
int x = c - '0'; // 1

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3. Unsafe Conversion (Common Mistake)

char c = char(1);

❌ Problem:

• Stores raw value 1
• Equivalent to '\x01' (non-printable)

✔ Not the same as

char c = '1'; // value = 49

Why + '0' is needed

char(bit + '0')

Because:

• bit = 0 → 0 + 48 = '0'
• bit = 1 → 1 + 48 = '1'

Binary Operations Case

ans = char(abit ^ bbit ^ carry) + ans; // ❌

• Produces '\x00' or '\x01'
• Not printable

ans = char((abit ^ bbit ^ carry) + '0') + ans;  // ✅

• Produces '0' or '1'

Summary Rules

Always use:

int → char  :  + '0'
char → int  :  - '0'

❌ Never assume:

char(1) == '1'  // WRONG

✔ Instead:

char(1) == '\x01'

'0' is the base

• Digits are offsets from ‘0’

'0' + 0 → '0'
'0' + 1 → '1'
'0' + 2 → '2'