Need help with this, It's C Programming. ### 3.8.1: If-else statement: Fix errors.#### InstructionRe-type the code and fix any errors. The code should convert non-positive numbers to 1.```cif (userNum > 0) printf("Positive.\n");else printf("Non-positive, converting to 1.\n"); userNum = 1;printf("Final: %d\n", userNum);```### Example Code with Syntax Errors```c#include int main(void) { int userNum; scanf("%d", &userNum); if(userNum > 0) printf("Positive.\n"); else printf("Non-positive, converting to 1.\n"); userNum = 1; printf("Final:%d\n", userNum); return 0;}```### ExplanationThe above code has a logical and syntactic error. The `else` statement will always execute because the `userNum = 1;` line is not within the scope of the `else` block. To correct it, place braces `{}` around the `else` block.### Correct Code```c#include int main(void) { int userNum; scanf("%d", &userNum); if(userNum > 0) { printf("Positive.\n"); } else { printf("Non-positive, converting to 1.\n"); userNum = 1; } printf("Final:%d\n", userNum); return 0;}```### Error Messages- **Failed to Compile:** ``` main.c:17:4: error: expected identifier or '(' before 'return' 17 | return 0; | ^~~~~~ main.c:18:1: error: expected identifier or '(' before '}' token 18 | } | ^ ``` - **Explanation:** The errors reported here suggest a missing syntax that should be addressed by using braces properly around the if-else statements. ### Additional Note:Although the reported line number is in the uneditable part of the code, the error actually exists in your code. Tools often highlight errors at the end of the code block when the true issue lies within misused statements or blocks.

Algorithm:Start the program.Read an integer input from the user and store it in the variable…

CHALLENGE 3.8.1: If-else statement: Fix errors. ACTIVITY Re-type the code and fix any errors. The code should convert non-positive numbers to 1. if (userNum > 0) printf("Positive.\n"); else printf("Non-positive, converting to 1.\n"); userNum = 1; printf("Final: %d\n", userNum); Learn how our autograder works 1 #include 3 int main(void) { int userfum; 5 6 scanf("%d", &userNum); if(userNum>0) printf("postive.\n"); 10 11 printf("Non-positive, converting to 1.\n"); 12 13 14 7 8 15} 16 17 18) else( userNum-1; } printf("final:%d\n", userfum); return 0; Run Failed to compile main.c:17:4: error: expected identifier or before 'return' 17 1 return 0; I main.c:18:1: error: expected identifier or (before ''token 18 ) Mete Although

CHALLENGE 3.8.1: If-else statement: Fix errors. ACTIVITY Re-type the code and fix any errors. The code should convert non-positive numbers to 1. if (userNum > 0) printf("Positive.\n"); else printf("Non-positive, converting to 1.\n"); userNum = 1; printf("Final: %d\n", userNum); Learn how our autograder works 1 #include 3 int main(void) { int userfum; 5 6 scanf("%d", &userNum); if(userNum>0) printf("postive.\n"); 10 11 printf("Non-positive, converting to 1.\n"); 12 13 14 7 8 15} 16 17 18) else( userNum-1; } printf("final:%d\n", userfum); return 0; Run Failed to compile main.c:17:4: error: expected identifier or before 'return' 17 1 return 0; I main.c:18:1: error: expected identifier or (before ''token 18 ) Mete Although

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Concept explainers

Operations

In mathematics and computer science, an operation is an event that is carried out to satisfy a given task. Basic operations of a computer system are input, processing, output, storage, and control.

Basic Operators

An operator is a symbol that indicates an operation to be performed. We are familiar with operators in mathematics; operators used in computer programming are—in many ways—similar to mathematical operators.

Division Operator

We all learnt about division—and the division operator—in school. You probably know of both these symbols as representing division:

Modulus Operator

Modulus can be represented either as (mod or modulo) in computing operation. Modulus comes under arithmetic operations. Any number or variable which produces absolute value is modulus functionality. Magnitude of any function is totally changed by modulo operator as it changes even negative value to positive.

Operators

In the realm of programming, operators refer to the symbols that perform some function. They are tasked with instructing the compiler on the type of action that needs to be performed on the values passed as operands. Operators can be used in mathematical formulas and equations. In programming languages like Python, C, and Java, a variety of operators are defined.

Question

100%

Need help with this, It's C Programming.

### 3.8.1: If-else statement: Fix errors.

#### Instruction
Re-type the code and fix any errors. The code should convert non-positive numbers to 1.

```c
if (userNum > 0)
printf("Positive.\n");
else
printf("Non-positive, converting to 1.\n");
userNum = 1;

printf("Final: %d\n", userNum);
```

### Example Code with Syntax Errors

```c
#include <stdio.h>

int main(void) {
int userNum;

scanf("%d", &userNum);

if(userNum > 0)
printf("Positive.\n");
else
printf("Non-positive, converting to 1.\n");
userNum = 1;

printf("Final:%d\n", userNum);

return 0;
}
```

### Explanation
The above code has a logical and syntactic error. The `else` statement will always execute because the `userNum = 1;` line is not within the scope of the `else` block. To correct it, place braces `{}` around the `else` block.

### Correct Code

```c
#include <stdio.h>

int main(void) {
int userNum;

scanf("%d", &userNum);

if(userNum > 0) {
printf("Positive.\n");
} else {
printf("Non-positive, converting to 1.\n");
userNum = 1;
}

printf("Final:%d\n", userNum);

return 0;
}
```

### Error Messages
- **Failed to Compile:**
```
main.c:17:4: error: expected identifier or '(' before 'return'
17 | return 0;
| ^~~~~~
main.c:18:1: error: expected identifier or '(' before '}' token
18 | }
| ^
```
- **Explanation:**
The errors reported here suggest a missing syntax that should be addressed by using braces properly around the if-else statements.

### Additional Note:
Although the reported line number is in the uneditable part of the code, the error actually exists in your code. Tools often highlight errors at the end of the code block when the true issue lies within misused statements or blocks.

Process by which instructions are given to a computer, software program, or application using code.

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