Data Type Sizes

[SercoSteveB]

Following execution of the VI which of the files created is the largest.

NOTE:  Although the VI Snippet hasn't been created properly, the Cluster used in the top file write is supposed to contain 3 DBL numerics with values 1,2 and 3.

a) steve1.bin

b) steve2.bin

c) steve3.bin

d) steve4.bin

[crossrulz]

Without running it, I would say B.

[Asha_Nagaraj]

Ans is B

[GayatriT]

With running it, I would say B.'

Array index will take 4 bytes, each element will take 8 bytes so here 3 elements are enabled so 8*3=24 bytes.  24+4=28 bytes

steve1.bin will take 4 bytes

steve2.bin will take 7 bytes

steve4.bin will take 8 bytes

[faymia]

GayatriT says it all... The answer is B with 28 bytes.

[crossrulz]

Except steve1.bin is actually 24 bytes.  It doesn't show in the picture, but the cluster actually has 3 doubles in it.

[RAMESHB]

Ans  is B

[swetha_sadula]

steve 2: 64*3=192 bits

steve 3: 8*3=24bits

steve 4:8*3+4=28bits

so answer:B

anybody will explain about option A?

[crossrulz]

setha_sadula,

It is easier to do the math with bytes instead of bits.  Everything works in bytes anyways.  GayatriT states how it all goes down (except for not knowing there are 3 doubles in the cluster).

The cluster is just flat.  It will save the 3 doubles as normal doubles in the file.  So steve1.bin will be 24 bytes.

The gotcha in this question is that when saving arrays to a binary file, by default the length of the array is saved before the array.  That is an extra I32 (4 bytes).  So steve2.bin will actually be 28 bytes.

Each character in a string is 1 byte.  So a 3 character string (steve3.bin) will be 3 bytes plus the length prepended (4 bytes), making 7 bytes total).

The final option is just saving a double (8 bytes).  So steve4.bin will be 8 bytes.

[TG2]

I have a question, I think this code does not run anyway. The Scan Value has 2 recommended input and only one is wired.

Ignore this, my answer is B

[ashwinilele]

B

[skian]

B