设置了两个mov ,出现以下问题。 3 }$ I- n% H% C2 P. v0 T+ F3 YYou lose, fella. The EMTP logic has detected an error condition, and is now going to terminate program execution. The following$ B. ?: @; |3 C$ T1 q
3 _4 E$ p+ Z+ c6 |9 j6 b5 J7 w8 ?. n
message summarizes the circumstances leading to this situation. Where an otherwise-unidentified data card is referred to, or where : D, Y+ h6 I; |1 ~5 ]) N8 W- n6 uthe "last" card is mentioned, it is the most recently read card of the input data that is meant. The 80-column image of this card 0 B$ W% I4 d. C" b) b _0 N* @is generally the last one printed out prior to this termination message. But possibly this last-read card has not yet been . G ]6 z' C& i8 Edisplayed, so a copy follows:! Q* |/ j8 E6 H9 [2 f1 S( R
" " : }* S) m( ]# V+ p' { KILL code number Overlay number Nearby statement number6 R1 G+ B4 Y# Y5 Y1 ^0 r
209 18 3501 / s3 l+ |. t$ p2 ]3 MKILL = 209. ZnO solution by Newton`s method of 2 coupled arresters has failed due to singularity of the Jacobian matrix. This9 {7 o. c0 x N: f7 [
is measured by tolerance EPSILN, which is defined by the floating-point miscellaneous data card. The iteration count is 2, and - I2 V/ D- d( p- u9 J' J! u. |the coupled group of arresters can be identified from the knowledge that the first (in order of input) connects bus "XX0035" with; ?& y* k/ D I+ h
bus "XX0010". The simulation time is T = 1.00000000E-06.. g6 p! j2 y1 [
9 B6 z8 c% ?0 R6 O7 A$ P