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* Convert the following C code to MIPS #include #include int main() { } int num1 = 23; int num2 = 24; int num3= 25; int num4 26; int num5 = 0; int arr[] = {10,11,13,14,15,16); = num5 = numl num2 arr [5] return 0; (num3 + num4); = arr [0] + arr [2];

* Convert the following C code to MIPS #include #include int main() { } int num1 = 23; int num2 = 24; int num3= 25; int num4 26; int num5 = 0; int arr[] = {10,11,13,14,15,16); = num5 = numl num2 arr [5] return 0; (num3 + num4); = arr [0] + arr [2];

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Hello. Please answer the attached MIPS programming question correctly by converting the given C code to MIPS. Please do not use very advanced syntax to solve the problem. 

*If correctly convert the code and do not use very advanced syntax, I will give you a thumbs up. Thanks.

* Convert the following C code to MIPS #include <stdio.h> #include <stdlib.h> int main() { } int numl = = 23; int num2 = 24; int num3 = 25; int num4 = 26; int num5 = 0; int arr[] = {10,11,13,14,15,16}; num5 = numl + num2 - (num3 + num4); arr [5] = arr [0] + arr [2]; return 0;
Transcribed Image Text:* Convert the following C code to MIPS #include <stdio.h> #include <stdlib.h> int main() { } int numl = = 23; int num2 = 24; int num3 = 25; int num4 = 26; int num5 = 0; int arr[] = {10,11,13,14,15,16}; num5 = numl + num2 - (num3 + num4); arr [5] = arr [0] + arr [2]; return 0;
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