Script - Bitcoin Programming Discussion

Bitcoin Programming 101
#201

Assignment Solution:

3 3 OP_MUL 2 2 OP_MUL OP_ADD
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#202

2 2 op_mul 3 3 op_mul op_add

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#203

Assignment Answer:

  1. 2 2 OP_MUL
  2. 3 3 OP_MUL
  3. OP_ADD

*I understood the assignment, but Iā€™m a little confused why BitcoinIDE shows the execution 0xD, shouldnā€™t it show 13?

#204

0xD is 13 in HEX. :slight_smile:

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#205

Thankyou! https://www.wikihow.com/Understand-Hexadecimal
Hereā€™s a link for anyone thatā€™s doesnā€™t understand HEX, the video helped me. :sweat_smile:

4 Likes
#206

2 2 op_mul 3 3 op_mul op_add

3 Likes
#207

The answer ist the following.

2 2 OP_MUL 3 3 OP_MUL OP_ADD

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#208

The code for [(2x2)+(3x3)] is as follows:
2 2 op_mul 3 3 op_mul op_add

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#209

2 2 op_mul 3 3 op_mul op_add

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#210

Answer:
2 2 OP_MUL 3 3 OP_MUL OP_ADD

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#211

Answer: 3 3 OP_MUL 2 2 OP_MUL OP_ADD

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#212

Answer
2 2 OP_MUL 3 3 OP_ADD OP_ ADD

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#213

2 2 op_mult 3 3 op_mult op_add

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#214

Answer
2 2 OP_MUL 3 3 OP_MUL OP_ADD

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#215

(2x2) + (3x3) = 2 2 OP_MUL 3 3 OP_MUL OP_ADD

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#216

Assignment exercise:
2 2 op_mul 3 3 op_mul op_add

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#217

Assignment Answer ! Bitcoin IDE :

2 2 OP_MUL 3 3 OP_MUL op_ADD

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#218

Answer : 2 2 OP_MUL
3 3 OP_MUL OP_ADD

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#219

Assignment Answer

2 2 OP_MUL 3 OP_ADD 3 OP_ MUL

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#220

This would not give the correct result. The program would execute 2 * 2 (= 4), then add 3 which would result in 7 and then multiply with 3. The formula would end up being (2 * 2 + 3) * 3.

You are close, try again :wink:

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