Okay so, I have uploaded my entire lab assignment ( just for clarity sake ). I have completed Part I and I would say half of Part II, but I am having trouble communicating two files to each other. I am already doing it once, but I don’t fully understand how I even did it so I don’t know how to apply it to another file. A quick summary of what the lab is, is a 3×3 maze (I will input what it looks like at the end but it may look weird with formatting) but I am to have a 3×3 maze with a 1 representing a wall, 2 represents the initial location, and 3 represents the end location. 002 010 030 This would be the Part 1 on the lab, but we are to replace wherever the player is (Number 2) with X, so it would be: 00X 010 030 And now I need to code instructions that I am able to move the X to the 3 and notify the user when you are unable to go a certain way due to a wall or out of bounds. Hopefully someone can help because the teacher was even willing to extend the deadline for me, but the book isn’t really helping with this.

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Hi this is the second time i am posting this and I am hoping that someone can actually help me! This is actually really simple for someone who knows how to code in Java!!! The code that is already given, CAN NOT BE ALTERED!!!!!! It is simply filling in the two spots where it says “enter code here”. I know the second part has a loop within a loop. If someone can please hep. Thanks! Remember, you CAN NOT ALTER the given code!

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Hi, I have been problems with this lab, please see attachment!

Note: You will use MultiSIM in XenDesktop to complete this assignment.

• Lab 5:555 Clock Timer and DC Motor Speed Control

Instructions for completing the Lab report

• The report is to be typed and prepared following the guidelines listed below.
• Send MultiSim Circuit Files or other software specific files used in the course (if applicable) separately from the lab report.

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A less-used fifth operation, SEEK(), allows you to set the read/write position directly without reading or writing.

Almost every programming language supports a version of this interface. You may recognize it from Python. For the C programmer, this interface is provided by these four system calls defined in stdio.h:

FILE * fopen( const char * filename, const char * mode);size_t fwrite( const void * ptr, size_t size, size_t nitems, FILE * stream);size_t fread( void * ptr,size_t size, size_t nitems, FILE * stream);int fclose( FILE *stream);Useful extras

You may find these useful:

• fseek() : repositions the current read/write location.
• feof() : tells you if the end-of-file is reached.
• ftell() : returns the current read/write location.
• ftruncate() : truncate a file to a specified length.
• stat() : get file status

Files by example

Examples of using the file API as demonstrated in class, and beyond. Background on files and links to the interface specifications are provided below.

Write a simple array to a file#include <stdio.h>int main( int argc, char* argv[] ){ const size_t len = 100; int arr[len]; // put data in the array // … // write the array into a file (error checks ommitted) FILE* f = fopen( “myfile”, “w” ); fwrite( arr, sizeof(int), len, f ); fclose( f ); return 0;}Read a simple array from a file#include <stdio.h>int main( int argc, char* argv[] ){ const size_t len = 100; int arr[len]; // read the array from a file (error checks ommitted) FILE* f = fopen( “myfile”, “r” ); fread( arr, sizeof(int), len, f ); fclose( f ); // use the array // … return 0;}Write an array of structs to a file, then read it back#include <stdio.h>#include <stdlib.h>#include <string.h>typedef struct { int x,y,z;} point3d_t;int main( int argc, char* argv[] ){ const size_t len = atoi(argv[1]); // array of points to write out point3d_t wpts[len]; // fill with random points for( size_t i=0; i<len; i++ ){wpts[i].x = rand() % 100;wpts[i].y = rand() % 100;wpts[i].z = rand() % 100;} // write the struct to a file (error checks ommitted) FILE* f1 = fopen( argv[2], “w” ); fwrite( wpts, sizeof(point3d_t), len, f1 ); fclose( f1 ); // array of points to read in from the same file point3d_t rpts[len]; // read the array from a file (error checks ommitted) FILE* f2 = fopen( argv[2], “r” ); fread( rpts, sizeof(point3d_t), len, f2 ); fclose( f2 ); if( memcmp( wpts, rpts, len * sizeof(rpts[0]) ) != 0 )puts( “Arrays differ” ); elseputs( “Arrays match” ); return 0;}Saving and loading an image structure, with error checking

This example shows the use of a simple file format that uses a short “header” to describe the file contents, so that an object of unknown size can be loaded.

Make sure you understand this example in detail. It combines elements from the examples above into a simple but realistic implementation of a file format.

/* saves an image to the filesytem using the file format:[ cols | rows | pixels ]where:cols is a uint32_t indicating image widthrows is a uint32_t indicating image heightpixels is cols * rows of uint8_ts indicating pixel grey levels*/int img_save( const img_t* img, const char* filename ){ assert( img ); assert( img->data ); FILE* f = fopen( filename, “w” ); if( f == NULL ){puts( “Failed to open image file for writing” );return 1;} // write the image dimensions header uint32_t hdr[2]; hdr[0] = img->cols; hdr[1] = img->rows; if( fwrite( hdr, sizeof(uint32_t), 2, f ) != 2 ){puts( “Failed to write image header” );return 2; const size_t len = img->cols * img->rows; if( fwrite( img->data, sizeof(uint8_t), len, f ) != len ){puts( “Failed to write image pixels” );return 3; fclose( f ); return 0;}/* loads an img_t from the filesystem using the same format as img_save().Warning: any existing pixel data in img->data is not free()d.*/int img_load( img_t* img, const char* filename ){ assert( img ); FILE* f = fopen( filename, “r” ); if( f == NULL ){puts( “Failed to open image file for reading” );return 1;} // read the image dimensions header: uint32_t hdr[2]; if( fread( hdr, sizeof(uint32_t), 2, f ) != 2 ){puts( “Failed to read image header” );return 2; img->cols = hdr[0]; img->rows = hdr[1]; // helpful debug: // printf( “read header: %u cols %u rowsn”, // img->cols, img->rows ); // allocate array for pixels now we know the size const size_t len = img->cols * img->rows; img->data = malloc( len * sizeof(uint8_t) ); assert( img->data ); // read pixel data into the pixel array if( fread( img->data, sizeof(uint8_t), len, f ) != len ) { puts( “Failed to read image pixels” ); return 3; fclose( f ); return 0;}

Usage:

img_t img;img_load( &img, “before.img” );image_frobinate( img ); // manipulate the image somehowimg_save( &img, “after.img” );Task 1: Serialize an array of integers to a binary-format file

Extend the functionality of your integer array from Lab 5 to support saving and loading arrays from the filesystem in a binary format.

Fetch the header file “intarr.h”. It contains these new function declarations:

/* LAB 6 TASK 1 *//* Save the entire array ia into a file called ‘filename’ in a binary file format that can be loaded by intarr_load_binary(). Returns zero on success, or a non-zero error code on failure. Arrays of length 0 should produce an output file containing an empty array.*/int intarr_save_binary( intarr_t* ia, const char* filename );/* Load a new array from the file called ‘filename’, that was previously saved using intarr_save_binary(). Returns a pointer to a newly-allocated intarr_t on success, or NULL on failure.*/intarr_t* intarr_load_binary( const char* filename );/* LAB 6 TASK 2 *//* Save the entire array ia into a file called ‘filename’ in a JSON text file array file format that can be loaded by intarr_load_json(). Returns zero on success, or a non-zero error code on failure. Arrays of length 0 should produce an output file containing an empty array. The JSON output should be human-readable. Examples: The following line is a valid JSON array: [ 100, 200, 300 ] The following lines are a valid JSON array: [ 100, 200, 300 ]*/int intarr_save_json( intarr_t* ia, const char* filename );/* Load a new array from the file called ‘filename’, that was previously saved using intarr_save(). The file may contain an array of length 0. Returns a pointer to a newly-allocated intarr_t on success (even if that array has length 0), or NULL on failure.*/intarr_t* intarr_load_json( const char* filename );Submission

Commit the single file “t2.c” to your repo in the lab 6 directory.

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Lab 6: Taxonomy

ANSWER KEY

Pre-Lab Questions

1. Use the following classifications to determine which organism is least related out of the three. Explain your rationale. (1 pts)

The Eastern Newt is the least related organism out of the three. While all three are classified into the same domain, kingdom, phylum and class the Eastern Newt is in a different order than the American Green Tree Frog and the European Fire-Bellied Toad.

2. How has DNA sequencing affected the science of classifying organisms? (1 pts)

DNA sequencing has allowed for the comparison of genes at the molecular level as opposed to physical traits at the organism level. Physical traits can be misleading when classifying how related two organisms are. DNA sequencing can also trace relatedness through generations and more accurately assess how closely related two organisms are.

3. You are on vacation and see an organism that you do not recognize. Discuss what possible steps you can take to classify it. (1 pts)

The organism’s physical features can be used to compare it to known organisms. Some physiological features can even possibly be used to help classify it.

Cla

Experiment 1: Dichotomous Key Practice Level American Green Tree

Fro  g 

European Fire-

Bellied Toadinomial Name

 Table 3: Dichotomous Key Results

(2 pts each)

Post-Lab Questions

1. What do you notice about the options of each step as they go from number one up. (1 pt)

The options become more and more specific.

2. How does your answer from question one relate to the Linnaean classification system? (1 pts)

The dichotomous key options became more and more specific as they came closer to identifying the organism just like the classification system starts as a broad category (i.e, animal kingdom) and becomes more specific until a unique species is classified (i.e., species).

Experiment 2: Classification of Organisms

The flow chart questions will lead you to the correct classification of the organisms into their respective kingdoms. Table 2, shown above, has an error in your lab manual–sunflowers do not have motility. Most of you saw the discrepancy and went with the answer you got from the flow chart. For the blanks in the completed table (above), that’s because those answers are variable, and not necessary to identify that organism using the given flow chart. (3 pts)

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