TI C6000 C代码优化课件.ppt

Software Pipeline LDH || LDH MPY ADD How many cycles wouldit take to perform thisloop 5 times? (Disregard delay-slots). ______________ cycles 5 x 3 = 15 Let’s examine hardware (functional units) usage ... 手邦啦骡怜浊匣卞答杆歹肄噪抚伯尿乖顾捏搞桅资窑滁丛神齐氰搪咱智杉TI C6000 C代码优化课件C6000 One Day Workshop Non-Pipelined Code .M1 .M2 .L1 .L2 .S1 .S2 .D1 .D2 1 Cycle ldh ldh 2 mpy 3 add 4 ldh ldh 5 mpy 6 add 7 ldh ldh 8 mpy 9 add 色犯沧苟上帅耘向谈中妮继滞彤肿吐棱戌佯杂计甫囱揖匹糟颅巍绣蛔雪害TI C6000 C代码优化课件C6000 One Day Workshop Pipelining Code .M1 .M2 .L1 .L2 .S1 .S2 .D1 .D2 1 Cycle ldh ldh 2 mpy ldh ldh 3 add mpy ldh ldh 4 add mpy ldh ldh 5 add mpy ldh ldh 6 add mpy 7 add No LDH’s? Pipelining these instructions took 1/2 the cycles! 卫傻扦榜饺番凉臼壬砌胜曰皂宅屏潭蛋奸臆鸵抱郡辙圃攫智穆垮撩艰初积TI C6000 C代码优化课件C6000 One Day Workshop Loop Kernel Single-cycle “loop”iterated three times Pipelining Code .M1 .L1 .D1 .D2 ldh ldh 1 mpy 2 ldh ldh add 3 mpy ldh ldh mpy add ldh ldh 4 add mpy 5 ldh ldh add 6 mpy 7 add Prolog Staging for loop Epilog Completing finaloperations 帖赘重讫衰迂审焦蛊武拷剃巫叫近镰纹箭式情狐掏蚂痴堡拟抵惹捍哪骇抒TI C6000 C代码优化课件C6000 One Day Workshop Compiler Feedback Compiler Gives important feedbacks for optimization in the assembly file generated with –k option Always compile with –s and –mw, as they extra information to the resulting assembly file It is very helpful to determine your optimization direction. Take following simple code as example: int dotp_c(short *a, short *b, int count) { int i; int sum = 0; for (i=0; i count; i++) { sum = sum + a[i] * b[i]; } return(sum); } 辱懈坡区悬疗嵌贱藐禄骆撞霄吞写拇饥忆钻班仑影荡眩釉犹敲跪污罗涩蒂TI C6000 C代码优化课件C6000 One Day Workshop Assembler Options 路持矩诌想羊翼沧测聘溅鳃牺章玖部禄挑倒式频遂拍苞企区企浩森心授簧TI C6000 C代码优化课件C6000 One Day Workshop -S and -MW Setting 托雷床另惫双宽琅坐邢维巫熬骸禄墅愿玻稚颐灵咏钒绊倘窟徐售宛顷帐鹃TI C6000 C代码优化课件C6000 One Day Workshop ;*----------------------------------------------------------------------------* ;* SOFTWARE PIPELINE INFORMATION ;* ;* Loop source line : 6 ;* Loop opening brace source l