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The matrix of a (7,9)was given.The address of the first byte
of a (1,1)matrix=1258.It takes 4 bytes to store the
number,then calculate the address of the last byte of a
(5,8) matrix??

Answers were Sorted based on User's Feedback



The matrix of a (7,9)was given.The address of the first byte of a (1,1)matrix=1258.It takes 4 bytes..

Answer / nash

1434




STARTING ADDRESS IS 1258.
EACH ELEMENT IS OF 4 BYTES.
SO (5,8)=44 I.E. POSITION OF ELEMENT
44*4=176
1258+176=1434
DO REPLY.

Is This Answer Correct ?    24 Yes 10 No

The matrix of a (7,9)was given.The address of the first byte of a (1,1)matrix=1258.It takes 4 bytes..

Answer / jyothi

STARTING ADDRESS IS 1258.
FIRST ELEMENT ADDRESS FROM 1258 TO 1261.
LIKE THAT 5,8 MATRIX ADDRESS FROM 1430 TO 1433
SO LAST BYTE ADDRESS OF 5,8 IS 1433

Is This Answer Correct ?    13 Yes 2 No

The matrix of a (7,9)was given.The address of the first byte of a (1,1)matrix=1258.It takes 4 bytes..

Answer / snehal

(1,1)=1258
after tat for 1 it contain 8
for 2=9
for 3=9
for 4=9
for 5=8
total=43
43*4=172
n 1258+172=1430

Is This Answer Correct ?    8 Yes 3 No

The matrix of a (7,9)was given.The address of the first byte of a (1,1)matrix=1258.It takes 4 bytes..

Answer / rohit

to find starting byte for matrix a(5,1):
1258+(4*(9*4))-(1)=1401

then to find last byte for matrix a(5,8):

1401+(8*4)=1433

Is This Answer Correct ?    1 Yes 0 No

The matrix of a (7,9)was given.The address of the first byte of a (1,1)matrix=1258.It takes 4 bytes..

Answer / vijay

plz post clr question..?

The matrix of a (7,9)was given.The address of the first byte
of a (1,1)matrix=1258.It takes 4 bytes to store the
number,then calculate the address of the last byte of a
(5,8)element??

ans is 1433
In column address increments by (8x4+4)36=clmn-gap
In row address increments by 4=row-gap
So
add of (5,8) element (4 clmn-gaps & 7 row-gaps )
=1258+4x36+4x7=1430
add of last byte=[1430,1431,1432,1433]
ans 1433

Is This Answer Correct ?    0 Yes 0 No

The matrix of a (7,9)was given.The address of the first byte of a (1,1)matrix=1258.It takes 4 bytes..

Answer / shobha

1258+5*8*4bytes=1418

Is This Answer Correct ?    0 Yes 0 No

The matrix of a (7,9)was given.The address of the first byte of a (1,1)matrix=1258.It takes 4 bytes..

Answer / sona

Question is confusing we are unable to decide wheather (x,y)
is a matrix or a position in the matrix

Is This Answer Correct ?    0 Yes 1 No

The matrix of a (7,9)was given.The address of the first byte of a (1,1)matrix=1258.It takes 4 bytes..

Answer / manish panwar

1258-1261 1262-1265 1266-1269 1270-1273 1274-1277 1278-1281 1282-1285 1286-1289

1290-1293 1294-1297 1298-1301 1302-1305 1306-1309 1310-1313 1314-1317 1318-1321

1322-1325 1326-1329 1330-1333 1334-1337 1338-1341 1342-1345 1346-1349 1350-1353

1354-1357 1358-1361 1362-1365 1366-1369 1370-1373 1374-1377 1378-1381 1382-1385

1386-1389 1390-1393 1394-1397 1398-1401 1402-1405 1406-1409 1410-1413 1414-1417



(5,8) matrix

4 4 4 4 4 4 4 4 = (4*7)+3=31 1258+31 = 1289
4 4 4 4 4 4 4 4 = 4*8=32 1289+32 = 1321
4 4 4 4 4 4 4 4 = 4*8=32 1321+32 = 1353
4 4 4 4 4 4 4 4 = 4*8=32 1353+32 = 1385
4 4 4 4 4 4 4 4 = 4*8=32 1385+32 = 1417

(32*5)-1 = (160 -1)= 159
1258+159 = 1417


So answer 1410

Is This Answer Correct ?    2 Yes 8 No

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