2. Relational operators: – These operators are used for comparison. These are also called comparison operators.<\/strong><\/p>\n\n\n\n

>=\u2026\u2026\u2026\u2026\u2026\u2026\u2026..Greater than or equal to<\/p>\n\n\n\n

<=\u2026\u2026\u2026\u2026\u2026\u2026\u2026..Less than or equal to<\/p>\n\n\n\n

!=\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026Not Equal to<\/p>\n\n\n\n

3. Logical operators: – These operators are used to combine or negate conditions.<\/strong><\/p>\n\n\n\n

Symbol\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026Purpose<\/p>\n\n\n\n

&&\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.Logical AND<\/p>\n\n\n\n

||\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026Logical OR<\/p>\n\n\n\n

4. Assignment operator:- This operator is used to Assign value to a variable.<\/strong><\/p>\n\n\n\n

\n
5. Increment or Decrement operator:- It is used to increase or decrease the value of a variable by one. It can be used in two ways.<\/strong><\/p>\n\n\n\n
Pre-Increment \\ Decrement:- when the operator is placed before the operands.<\/li><\/ol>\n\n\n\n
 Example ++a, – -a;<\/p>\n\n\n\n
2.Post-Increment \\ Decrement:- when the operator is placed after the operand.<\/li><\/ul>\n\n\n\n
Example a++, a- –<\/p>\n<\/div><\/div>\n\n\n\n
\n
\n
Q. Write a program to represent pre-increment.<\/em><\/strong><\/p><\/blockquote>\n<\/div><\/div>\n<\/div><\/div>\n\n\n\n
#include<iostream.h>\n#include<conio.h>\nvoid main() \n{\t\nclrscr();\t\nint a,b;\na=5;\nb=++a;\t\ncout<<\u201da=\u201d<<a;\ncout<<\u201d\\n b=\u201d<<b;\ngetch();\n}\n<\/code><\/pre>\n\n\n\n\nOutput<\/p>\n\n\n\n a=6<\/p>\n\n\n\n b=6<\/p>\n\n\n\n <\/p>\n\n\n\n Q. Write a program to represent pre-decrement.<\/p><\/blockquote>\n\n\n\n <\/p>\n<\/div><\/div>\n\n\n\n #include<iostream.h>\n#include<conio.h>\nvoid main() \n{\nclrscr();\t\nint a,b;\na=5;\nb= - -a;\t\ncout<<\u201da=\u201d<<a;\ncout<<\u201d\\n b=\u201d<<b;\ngetch();\n}\n<\/code><\/pre>\n\n\n\nOutput<\/p>\n\n\n\n a=4<\/p>\n\n\n\n b=4<\/p>\n\n\n\n <\/p>\n\n\n\n Q. Write a program to represent post-increment.<\/p><\/blockquote>\n\n\n\n #include<iostream.h>\n#include<conio.h>\nvoid main() \n{\nclrscr();\t\nint a,b;\na=5;\nb=a++;\t\ncout<<\u201da=\u201d<<a;\ncout<<\u201d\\n b=\u201d<<b;\ngetch();\n}\n<\/code><\/pre>\n\n\n\nOutput<\/p>\n\n\n\n a=6<\/p>\n\n\n\n b=5<\/p>\n\n\n\n Q. Write a program to represent post-decrement.<\/em><\/strong><\/p><\/blockquote>\n\n\n\n #include<iostream.h>\n#include<conio.h>\nvoid main() \n{\nclrscr();\t\nint a,b;\na=5;\nb=a- -;\t\ncout<<\u201da=\u201d<<a;\ncout<<\u201d\\n b=\u201d<<b;\ngetch();\n}\n<\/code><\/pre>\n\n\n\nOutput<\/p>\n\n\n\n a=4<\/p>\n\n\n\n b=5<\/p>\n\n\n\n <\/p>\n\n\n\n Q. Find Output<\/strong><\/p><\/blockquote>\n\n\n\n <\/p>\n\n\n\n #include<iostream.h>\n#include<conio.h>\nvoid main() \n{\nclrscr();\t\nint a;\na=5;\n++a;\na++;\n++a;\na++;\ncout<<\u201da=\u201d<<a;\ngetch();\n}\n<\/code><\/pre>\n\n\n\nOutput<\/p>\n\n\n\n a=9<\/p>\n\n\n\n Q. Find Output<\/strong><\/p><\/blockquote>\n\n\n\n #include<iostream.h>\n#include<conio.h>\nvoid main() \n{\nclrscr();\t\nint a;\na=5;\n--a;\na--;\n--a;\na--;\ncout<<\u201da=\u201d<<a;\ngetch();\n}\n<\/code><\/pre>\n\n\n\nOutput<\/p>\n\n\n\n a=1<\/p>\n\n\n\n Q. Find Output<\/p><\/blockquote>\n\n\n\n #include<iostream.h>\n#include<conio.h>\nvoid main() \n{\nclrscr();\t\nint a;\na=5;\n++a;\na++;\n++a;\na--;\n--a;\ncout<<\u201da=\u201d<<a;\ngetch();\n}\n<\/code><\/pre>\n\n\n\nOutput<\/p>\n\n\n\n a=6<\/p>\n\n\n\n Q. Find Output<\/p><\/blockquote>\n\n\n\n #include<iostream.h>\n#include<conio.h>\nvoid main() \n{\nclrscr();\t\nint a;\na=5;\ncout<<++a;\ngetch();\n}\n<\/code><\/pre>\n\n\n\nOutput<\/p>\n\n\n\n 6<\/p>\n\n\n\n Q. Find Output<\/p><\/blockquote>\n\n\n\n #include<iostream.h>\n#include<conio.h>\nvoid main() \n{\nclrscr();\t\nint a;\na=5;\ncout<<- -a;\ngetch();\n}\n<\/code><\/pre>\n\n\n\nOutput<\/p>\n\n\n\n 4<\/p>\n\n\n\n Q. Find Output<\/p><\/blockquote>\n\n\n\n #include<iostream.h>\n#include<conio.h>\nvoid main() \n{\nclrscr();\t\nint a;\na=5;\ncout<<++a<<++a<<++a;\ngetch();\n}\n<\/code><\/pre>\n\n\n\nOutput<\/p>\n\n\n\n 8 7 6<\/p>\n\n\n\n Q. Find Output<\/p><\/blockquote>\n\n\n\n #include<iostream.h>\n#include<conio.h>\nvoid main() \n{\nclrscr();\t\nint a;\na=5;\ncout<<- -a<<- -a<<- -a;\ngetch();\n}\n<\/code><\/pre>\n\n\n\nOutput<\/p>\n\n\n\n 2 3 4<\/p>\n\n\n\n Q. Find Output<\/p><\/blockquote>\n\n\n\n #include<iostream.h>\n#include<conio.h>\nvoid main() \n{\nclrscr();\t\nint a;\na=5;\ncout<<a++<<a++<<a++;\ngetch();\n}\n<\/code><\/pre>\n\n\n\nOutput<\/p>\n\n\n\n 7 6 5<\/p>\n\n\n\n Q. Find Output<\/p><\/blockquote>\n\n\n\n #include<iostream.h>\n#include<conio.h>\nvoid main() \n{\nclrscr();\t\nint a;\na=5;\ncout<<a- -<<a- -<<a- -;\ngetch();\n}\n<\/code><\/pre>\n\n\n\nOutput<\/p>\n\n\n\n 3 4 5<\/p>\n\n\n\n Q. Find Output<\/p><\/blockquote>\n\n\n\n #include<iostream.h>\n#include<conio.h>\nvoid main() \n{\nclrscr();\t\nint a;\na=5;\ncout<<a++<<++a<<a++;\ngetch();\n}\n<\/code><\/pre>\n\n\n\nOutput<\/p>\n\n\n\n 7 7 5<\/p>\n\n\n\n Q. Find Output<\/p><\/blockquote>\n\n\n\n #include<iostream.h>\n#include<conio.h>\nvoid main() \n{\nclrscr();\t\nint a;\na=5;\ncout<<a- -<<- -a<<a- -;\ngetch();\n}\n<\/code><\/pre>\n\n\n\nOutput<\/p>\n\n\n\n 3 3 5<\/p>\n\n\n\n \nQ. Evaluate the following C++ expression s where a,b,c <\/strong>are integers and d,f<\/strong> are floating point numbers .The values are a=5,b=6<\/strong> and d=3.5<\/strong>.<\/p>\n\n\n\n a) c=a++ – (b ++) * (- -d)<\/strong><\/p>\n\n\n\n b) f=(++b) * b \u2013 a++<\/strong><\/p>\n<\/div><\/div>\n\n\n\n <\/p>\n\n\n\n \nQ. Evaluate the following C++ expression s where a,b,c <\/strong>are integers and d,f<\/strong> are floating point numbers .The values are a=3,b=5<\/strong> and d=1.5<\/strong>.<\/p>\n\n\n\n a) c=a++ +b*++d<\/strong><\/p>\n\n\n\n b) f= b*b++ – ++a<\/strong><\/p>\n<\/div><\/div>\n\n\n\n <\/p>\n\n\n\n Write the output of the following program:<\/p><\/blockquote>\n\n\n\n #include<iostream.h>\nvoid main()\n{\nint x=5,y=5;\ncout<<x++;\ncout<<\u201d,\u201d;\ncout<<++x;\ncout<<\u201d,\u201d;\ncout<<y++<<\u201d,\u201d<<++y;\n}\n<\/code><\/pre>\n\n\n\n<\/p>\n\n\n\n \nQ. Evaluate the following C++ expression s where a,b,c <\/strong>are integers and d,f<\/strong> are floating point numbers .The values are a=4,b=4<\/strong> and d=1.5<\/strong>.<\/p>\n\n\n\n a) c=a+++++b*++d<\/strong><\/p>\n\n\n\n b) f=++ b*b++ – ++a<\/strong><\/strong><\/p>\n<\/div><\/div>\n\n\n\n <\/p>\n\n\n\n Q. For a given variable \u201ca=, initialized by 10, what is the difference between statements a++ and ++a? Explain briefly<\/strong><\/p>\n\n\n\n <\/p>\n\n\n\n Bitwise operators:- These are used for bit manipulation. it can be classified into two groups.<\/strong><\/p>\n\n\n\n Bitwise Logical<\/li> Bitwise shift<\/li><\/ol>\n\n\n\n\n&………………………..Bitwise AND<\/p>\n\n\n\n |…………………………Bitwise OR<\/p>\n\n\n\n ~ ………………………..Bitwise NOT<\/p>\n\n\n\n ^…………………………Bitwise XOR<\/p>\n<\/div><\/div>\n\n\n\n \n<< ………………………Left shift<\/p>\n\n\n\n >>……………………….Right shift<\/p>\n<\/div><\/div>\n\n\n\n <\/p>\n\n\n\n \nBitwise AND: – This operator returns 1 when both bits are 1 otherwise it returns 0.<\/strong><\/p>\n\n\n\n \nA…………………B…………………….A&B<\/p>\n\n\n\n 0…………………0……………………0<\/p>\n\n\n\n 0………………….1…………………..0<\/p>\n\n\n\n 1…………………..0…………………0<\/p>\n\n\n\n 1…………………..1………………….1<\/p>\n<\/div><\/div>\n<\/div><\/div>\n\n\n\n Rule:- <\/p>\n\n\n\n Find binary of First variable.<\/li> Find binary of Second variable.<\/li> compare binary of both variable according to above rule.<\/li> Find decimal of what we get from comparison.<\/li><\/ul>\n\n\n\n<\/p>\n\n\n\n Q.Write a program to represent Bitwise AND.<\/p><\/blockquote>\n\n\n\n #include<iostream.h>\n#include<conio.h>\nvoid main() \n{\nclrscr();\t\nint a,b,c;\na=10;\nb=5;\nc=a&b;\ncout<<c;\ngetch();\n}\n<\/code><\/pre>\n\n\n\nOutput<\/p>\n\n\n\n 0<\/p>\n\n\n\n \nBitwise OR: \u2013 This operator returns 0 when both bits are 0 otherwise it returns 1.<\/strong><\/p>\n\n\n\n A\u2026\u2026\u2026\u2026\u2026\u2026\u2026B\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.A|B<\/p>\n\n\n\n 0\u2026\u2026\u2026\u2026\u2026\u2026\u20260\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u20260<\/p>\n\n\n\n 0\u2026\u2026\u2026\u2026\u2026\u2026\u2026.1\u2026\u2026\u2026\u2026\u2026\u2026\u2026..1<\/p>\n\n\n\n 1\u2026\u2026\u2026\u2026\u2026\u2026\u2026..0\u2026\u2026\u2026\u2026\u2026\u2026\u20261<\/p>\n\n\n\n 1\u2026\u2026\u2026\u2026\u2026\u2026\u2026..1\u2026\u2026\u2026\u2026\u2026\u2026\u2026.1<\/p>\n<\/div><\/div>\n\n\n\n Rule:-<\/p>\n\n\n\n Find binary of First variable.<\/li> Find binary of Second variable.<\/li> compare binary of both variable according to above rule.<\/li> Find decimal of what we get from comparison.<\/li><\/ul>\n\n\n\nQ.Write a program to represent Bitwise OR.<\/p><\/blockquote>\n\n\n\n #include<iostream.h>\n#include<conio.h>\nvoid main() \n{\nclrscr();\t\nint a,b,c;\na=10;\nb=5;\nc=a|b;\ncout<<c;\ngetch();\n}\n<\/code><\/pre>\n\n\n\nOutput<\/strong><\/p>\n\n\n\n 15<\/p>\n\n\n\n \nBitwise XOR: \u2013 This operator returns 0 when both bits are same otherwise it returns 1.<\/strong><\/p>\n\n\n\n A\u2026\u2026\u2026\u2026\u2026\u2026\u2026B\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.A^B<\/p>\n\n\n\n 0\u2026\u2026\u2026\u2026\u2026\u2026\u20260\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u20260<\/p>\n\n\n\n 0\u2026\u2026\u2026\u2026\u2026\u2026\u2026.1\u2026\u2026\u2026\u2026\u2026\u2026\u2026..1<\/p>\n\n\n\n 1\u2026\u2026\u2026\u2026\u2026\u2026\u2026..0\u2026\u2026\u2026\u2026\u2026\u2026\u20261<\/p>\n\n\n\n 1\u2026\u2026\u2026\u2026\u2026\u2026\u2026..1\u2026\u2026\u2026\u2026\u2026\u2026\u2026.0<\/p>\n<\/div><\/div>\n\n\n\n Rule:-<\/p>\n\n\n\n Find binary of First variable.<\/li> Find binary of Second variable.<\/li> compare binary of both variable according to above rule.<\/li> Find decimal of what we get from comparison.<\/li><\/ul>\n\n\n\nQ. Write a program to represent Bitwise XOR.<\/p><\/blockquote>\n\n\n\n #include<iostream.h>\n#include<conio.h>\nvoid main() \n{\nclrscr();\t\nint a,b,c;\na=10;\nb=5;\nc=a^b;\ncout<<c;\ngetch();\n}\n<\/code><\/pre>\n\n\n\nOutput<\/p>\n\n\n\n 15<\/p>\n\n\n\n \nBitwise NOT:- It inverts 0 to 1 and vice versa<\/p>\n\n\n\n A………………….~A<\/p>\n\n\n\n 0……………………..1<\/p>\n\n\n\n 1………………………0<\/p>\n<\/div><\/div>\n\n\n\n Rule:-<\/p>\n\n\n\n Find binary of variable.<\/li> change according to above rule.<\/li> Find decimal of what we get form change.<\/li><\/ul>\n\n\n\nQ. Write a program to represent Bitwise NOT.<\/p><\/blockquote>\n\n\n\n #include<iostream.h>\n#include<conio.h>\nvoid main() \n{\nclrscr();\t\nint a,b;\na=5;\nb=~a;\ncout<<b;\ngetch();\n}\n<\/code><\/pre>\n\n\n\nOutput<\/p>\n\n\n\n -6<\/p>\n\n\n\n Left Shift Rule <\/strong><\/p>\n\n\n\n Find binary of given variable.<\/li> Add zero ‘how many’ bits to be shifted.<\/li> write decimal of all bits.<\/li><\/ul>\n\n\n\nQ. Write a program to represent Left shift.<\/strong><\/p><\/blockquote>\n\n\n\n #include<iostream.h>\n#include<conio.h>\nvoid main() \n{\nclrscr();\t\nint a,b;\na=5;\nb=a<<2;\ncout<<b;\ngetch();\n}\n<\/code><\/pre>\n\n\n\nOutput<\/p>\n\n\n\n 20<\/p>\n\n\n\n Right Shift Rule<\/strong><\/p>\n\n\n\n Find binary of given variable.<\/li> Remove zero \u2018how many\u2019 bits to be shifted.<\/li> write decimal of Remaining bits.<\/li><\/ul>\n\n\n\nQ. Write a program to represent Right shift.<\/p><\/blockquote>\n\n\n\n #include<iostream.h>\n#include<conio.h>\nvoid main() \n{\nclrscr();\t\nint a,b;\na=5;\nb=a>>2;\ncout<<b;\ngetch();\n}\n<\/code><\/pre>\n\n\n\nOutput<\/p>\n\n\n\n 1<\/p>\n\n\n\n Special operators:- <\/strong>These operators are used for special purpose.<\/p>\n\n\n\n sizeof( ) <\/strong><\/p>\n\n\n\n This operator is used to find size of data type or variable.<\/p>\n\n\n\n Syntax <\/p>\n\n\n\n sizeof(datatype \/ Variable);<\/strong><\/p>\n\n\n\n Q. Write a program to represent sizeof( ) operator.<\/p><\/blockquote>\n\n\n\n #include<iostream.h>\n#include<conio.h>\nvoid main() \n{\nclrscr();\t\ncout<<sizeof(int);\ngetch();\n}\n<\/code><\/pre>\n\n\n\nOutput<\/p>\n\n\n\n 2<\/p>\n\n\n\n OR<\/strong><\/p>\n\n\n\n #include<iostream.h>\n#include<conio.h>\nvoid main() \n{\nclrscr();\t\nint z;\nz= sizeof(int);\ncout<<\u201dsize of integer=\u201d<<z;\ngetch();\n}\n<\/code><\/pre>\n\n\n\nOutput <\/p>\n\n\n\n size of integer= 2<\/p>\n\n\n\n Conditional or Ternary operator<\/strong>:-<\/p>\n\n\n\n It is a conditional statement and used for making decision when we have two options. <\/strong><\/p>\n\n\n\n It requires three operands so , also known as ternary operator.<\/p>\n\n\n\n syntax<\/p>\n\n\n\n condition ? <true block> : <false block> ;<\/strong><\/p>\n\n\n\n <\/p>\n\n\n\n <\/p>\n\n\n\n <\/p>\n\n\n\n Comma(,) <\/strong><\/p>\n\n\n\n This operator is used for variable separator.<\/strong><\/p>\n\n\n\n Q. write a program to check person is eligible for vote or not. using Ternary or conditional operator.compile time<\/p><\/blockquote>\n\n\n\n #include<iostream.h>\n#include<conio.h>\nvoid main()\n{\nclrscr();\nint a;\na=10;\na>=18?cout<<\u201d eligible for vote\u201d:cout<<\u201d not eligible for vote\u201d;\ngetch();\n}\n<\/code><\/pre>\n\n\n\noutput<\/strong><\/p>\n\n\n\n not eligible for vote<\/p>\n\n\n\n Q. write a program to check person is eligible for vote or not. (using Ternary or conditional operator.Run time.)<\/p><\/blockquote>\n\n\n\n #include<iostream.h>\n#include<conio.h>\nvoid main()\n{\nclrscr();\nint a;\ncout<<\u201dEnter age\u201d;\ncin>>a;\na>=18?cout<<\u201d eligible for vote\u201d:cout<<\u201d not eligible for vote\u201d;\ngetch();\n}\n<\/code><\/pre>\n\n\n\noutput<\/strong><\/p>\n\n\n\n Enter age=12<\/p>\n\n\n\n not eligible for vote<\/p>\n\n\n\n Q. write a program to check number is even or odd. (using Ternary or conditional operator.compile time.)<\/p><\/blockquote>\n\n\n\n #include<iostream.h>\n#include<conio.h>\nvoid main()\n{\nclrscr();\nint n;\nn=5;\nn%2==0?cout<<\u201d Even number\u201d:cout<<\u201dOdd number\u201d;\ngetch();\n}\n<\/code><\/pre>\n\n\n\noutput<\/strong><\/p>\n\n\n\n Odd number<\/p>\n\n\n\n Q. write a program to check number is even or odd. (using Ternary or conditional operator.Run time.)<\/p><\/blockquote>\n\n\n\n #include<iostream.h>\n#include<conio.h>\nvoid main()\n{\nclrscr();\nint n;\ncout<<\u201dEnter any number=\u201d;\ncin>>n;\nn%2==0?cout<<\u201d Even number\u201d:cout<<\u201dOdd number\u201d;\ngetch();\n}\n<\/code><\/pre>\n\n\n\noutput<\/strong><\/p>\n\n\n\n Enter any number=5<\/p>\n\n\n\n Odd number<\/p>\n\n\n\n <\/p>\n\n\n\n <\/p>\n\n\n\n <\/p>\n\n\n\n <\/p>\n\n\n\n <\/p>\n\n\n\n \n<\/iframe>\n<\/div><\/figure>\n\n\n\n<\/p>\n\n\n\n<\/p>\n\n\n\n<\/p>\n\n\n\n<a href=\"http:\/\/www.nipsacademy.com\/blogs\/c-programming-language-tutorial\/\">Also Read:- C++ Programming Tutorial<\/a><\/p>\n\n\n\n<a href=\"http:\/\/www.nipsacademy.com\/blogs\/computer-science-xi-jac-board-2020-answer-key\/\">Also Read:- computer Science Exam 2020<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"Operators are the symbol that represent operation between operands. C=A+B In above example A,B and C are the operands and +, = are the operators. Note:- Operand:- on which operation is to be performed. Expression:- combination of operator and operands. There are following types of operators in C++ language that perform different operation. Arithmetic operators […]<\/p>\n","protected":false},"author":1,"featured_media":323,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[4,5,3,9,7,8,6],"tags":[],"class_list":["post-322","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-11th-computer-science","category-12th-computer-science","category-bca","category-cbse-12th-computer-science","category-cbse","category-jac-12th-computer-science","category-jac"],"_links":{"self":[{"href":"https:\/\/aipsacademy.com\/blogs\/wp-json\/wp\/v2\/posts\/322","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/aipsacademy.com\/blogs\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/aipsacademy.com\/blogs\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/aipsacademy.com\/blogs\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/aipsacademy.com\/blogs\/wp-json\/wp\/v2\/comments?post=322"}],"version-history":[{"count":3,"href":"https:\/\/aipsacademy.com\/blogs\/wp-json\/wp\/v2\/posts\/322\/revisions"}],"predecessor-version":[{"id":327,"href":"https:\/\/aipsacademy.com\/blogs\/wp-json\/wp\/v2\/posts\/322\/revisions\/327"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/aipsacademy.com\/blogs\/wp-json\/wp\/v2\/media\/323"}],"wp:attachment":[{"href":"https:\/\/aipsacademy.com\/blogs\/wp-json\/wp\/v2\/media?parent=322"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/aipsacademy.com\/blogs\/wp-json\/wp\/v2\/categories?post=322"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/aipsacademy.com\/blogs\/wp-json\/wp\/v2\/tags?post=322"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}