Common string operations - Step-by-Step Execution

Concept Flow - Common string operations

Start

↓

Declare strings

↓

Perform operations

↓

Length calculation

↓

Copy string

↓

Concatenate strings

↓

Compare strings

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Output results

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End

This flow shows declaring strings, performing common operations like length, copy, concatenate, compare, then outputting results.

Execution Sample

C

#include <stdio.h>
#include <string.h>

int main() {
  char str1[20] = "Hello";
  char str2[] = "World";
  strcat(str1, str2);
  printf("%s", str1);
  return 0;
}

This code concatenates two strings and prints the result.

Execution Table

Step	Action	str1	str2	Output
1	Declare str1 as "Hello"	Hello	Uninitialized
2	Declare str2 as "World"	Hello	World
3	Concatenate str2 to str1 (strcat)	HelloWorld	World
4	Print str1	HelloWorld	World	HelloWorld
5	Return 0, program ends	HelloWorld	World

💡 Program ends after printing concatenated string.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	Final
str1	Uninitialized	Hello	Hello	HelloWorld	HelloWorld
str2	Uninitialized	Uninitialized	World	World	World

Key Moments - 3 Insights

Why does str1 need to have enough space before strcat?

Because strcat adds characters to str1, if str1's array is too small, it will overwrite memory causing errors. See step 3 where str1 changes from "Hello" to "HelloWorld".

Why can't we assign strings with = after declaration?

In C, you can only assign strings at declaration. After that, you must use functions like strcpy. This is why str1 is declared with "Hello" initially (step 1), not assigned later.

What happens if we print str2 after strcat?

str2 remains unchanged as "World" (step 3 and 4). strcat only modifies str1, not str2.

Visual Quiz - 3 Questions

Look at the execution table, what is the value of str1 after step 3?

A "World"

B "HelloWorld"

C "Hello"

D Uninitialized

💡 Hint

Check the 'str1' column in row for step 3 in the execution_table.

At which step is the output "HelloWorld" produced?

A Step 4

B Step 2

C Step 3

D Step 5

💡 Hint

Look at the 'Output' column in the execution_table.

If str1 was declared as char str1[5] = "Hello"; what would happen at step 3?

A str1 becomes empty string

B Concatenation succeeds safely

C Buffer overflow may occur

D str2 changes to "HelloWorld"

💡 Hint

Refer to key_moments about space needed for strcat.

Concept Snapshot

Common string operations in C:
- Use <string.h> functions
- strlen(str): get length
- strcpy(dest, src): copy string
- strcat(dest, src): append string
- strcmp(str1, str2): compare strings
- Arrays must have enough space for strcat
- Strings assigned only at declaration

Full Transcript

This visual trace shows common string operations in C. We start by declaring two strings, str1 and str2. Then we concatenate str2 to str1 using strcat. The variable str1 changes from "Hello" to "HelloWorld" after concatenation. We print str1, which outputs "HelloWorld". The program ends after printing. Important points: str1 must have enough space to hold the combined string, and strings cannot be assigned with = after declaration but must use functions like strcpy. str2 remains unchanged after concatenation. This helps beginners see how string operations change variables step-by-step.