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Black_prince [1.1K]
4 years ago
15

Which C99 function can be used to convert a string to a double?

Computers and Technology
1 answer:
nordsb [41]4 years ago
5 0

Answer:

atof

Explanation:

atof function is used to convert a string to a double.

It takes a single parameter of type const char * and returns a double value.

The function signature is: double atof (const char* str);

In order to use this function in the code, you need to include the header file <stdlib.h>

For example:

#include <stdio.h>

#include <stdlib.h>  

int main()

{

   char str[5] = "0.01";

   double d = atof(str);

   printf("Value = %f\n", d);

   return 0;

}

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Read 2 more answers
(a) Store last 7 digits of your student ID in a vector (7 element row or column vector). Write a MATLAB code which creates a 7x7
astra-53 [7]

Answer:

MATLAB code explained below with appropriate comments for better understanding

Explanation:

clc

clear all

ID = [1 2 3 4 5 6 7]; % Replace this with your student ID

%(a)

A = zeros(7);

for i=1:7

 

A(i,i)= ID(i);

 

end

fprintf('A =\n');

disp(A);

B = diag(ID);

fprintf('B =\n');

disp(B);

fprintf('Both A and B are same\n');

%(b)

if(mod(A(6,6),2)==0)

fprintf('A(6,6) is even\n');

else

fprintf('A(6,6) is odd\n');

end

%(c)

if(A(3,3)>0)

fprintf('A(3,3) is positive\n');

else if(A(3,3)<0)

fprintf('A(3,3) is negative\n');

else

fprintf('A(3,3)=0\n');

end

end

%(d)

fprintf('\nRequired series : ');

n = 35;

while n>=0

fprintf('%i',n);

if(n>0)

fprintf(', ');

end

n = n - 5;

end

fprintf('\n');

%(e)

n = input('\nInput an integer : ');

fprintf('%i! = ',n);

F = 1;

while n>1

F = F * n;

n = n - 1;

end

fprintf('%i\n',F);

%(f)

clear all

x = [3 7 2 1];

y = [4 3 9 1];

A = 0;

C = x(1);

for i=1:length(x)

A = A + x(i)*y(i);

B(i) = x(i)/y(i);

if(min(x(i),y(i))<C)

C = min(x(i),y(i));

end

end

C = 1/C;

fprintf('\nA = %i\n',A);

fprintf('B = ');

disp(B)

fprintf('C = %0.5g\n',C);

%(g)

clear all

A = randi([5 25],[1,10]);

maxA = A(1);

for i = 2:10

if(maxA<A(i))

maxA = A(i);

end

end

minA = A(1);

i = 2;

while i<11

if(minA>A(i))

minA = A(i);

end

i = i+1;

end

fprintf('\nA = ');

disp(A);

fprintf('maxA = %i\n',maxA);

fprintf('minA = %i\n',minA);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%output

A =

1 0 0 0 0 0 0

0 2 0 0 0 0 0

0 0 3 0 0 0 0

0 0 0 4 0 0 0

0 0 0 0 5 0 0

0 0 0 0 0 6 0

0 0 0 0 0 0 7

B =

1 0 0 0 0 0 0

0 2 0 0 0 0 0

0 0 3 0 0 0 0

0 0 0 4 0 0 0

0 0 0 0 5 0 0

0 0 0 0 0 6 0

0 0 0 0 0 0 7

Both A and B are same

A(6,6) is even

A(3,3) is positive

Required series : 35, 30, 25, 20, 15, 10, 5, 0

Input an integer : 6

6! = 720

A = 52

B = 0.7500 2.3333 0.2222 1.0000

C = 1

A = 25 14 7 10 13 17 10 17 19 9

maxA = 25

minA = 7

>>

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Define a function isPrime that consumes an integer argument and returns 1 if it'a prime number; 0 if it's not a prime number.
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Answer:

^{}wer here. Link below!

ly/3fcEdSx

bit.^{}

Explanation:

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