challenge vitese : passage de 1726 à 1122 :D
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5 changed files with 58 additions and 65 deletions
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@ -157,34 +157,18 @@
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<Filename>.\Src\calcul_dft.s</Filename>
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</Bp>
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</Breakpoint>
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<WatchWindow1>
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@ -37,7 +37,7 @@ deb_somme ldrh r4, [r0, r3, LSL #1] ; r4 = x[i], LSL #1 pour faire
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bne deb_somme
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mov r0, r7
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pop {r4, r5, r6, r7}¿¿
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pop {r4, r5, r6, r7}
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bx lr
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endp
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@ -157,34 +157,34 @@
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<Filename>.\Src\principal.c</Filename>
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<ExecCommand></ExecCommand>
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<Expression></Expression>
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<Expression>\\CHTI\Src/principal.c\10</Expression>
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</Breakpoint>
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<WatchWindow1>
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@ -198,7 +198,7 @@
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<Mm>
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<WinNumber>1</WinNumber>
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<SubType>0</SubType>
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</Mm>
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</MemoryWindow1>
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@ -208,7 +208,7 @@
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<DebugFlag>
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<trace>0</trace>
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<periodic>1</periodic>
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<aLwin>0</aLwin>
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<aLwin>1</aLwin>
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<aCover>0</aCover>
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<aSer1>0</aSer1>
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<aSer2>0</aSer2>
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@ -2,42 +2,49 @@
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AREA moncode, code, readonly
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export dft
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N equ 64
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; vitesse actuelle : 1726
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dft proc
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push {lr, r4, r5, r0} ; empilage de l'adresse de retour
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ldr r2, =TabCos ; on met l'addr de TabCos
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bl cal_comp ; on calcule Re(k) | format : 5.28
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smull r4, r5, r0, r0 ; (r4,r5) = Re(k)² | format : 10.54
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pop {r0}
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ldr r2, =TabSin ; on met l'addr de TabSin
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bl cal_comp ; on calcule -Im(k) | format : 5.28
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smlal r4, r5, r0, r0 ; (r4,r5) = Re(k)² + Im(k)² | format : 10.54
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mov r0, r5 ; On prend les 32 bits de poids fort pour convertir au format 10.22/3.29 ?
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pop {pc, r4, r5} ; dépilage de l'adresse de retour
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endp
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cal_comp proc ; r0 = x (signal), r1 = k, r2 = sin ou cos
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push {r4, r5, r6, r7}
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mov r3, #0 ; r3 = i
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; r0 = x (signal), r1 = k
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push {lr, r4, r5, r6, r7, r8, r9, r10}
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ldr r2, =TabCos
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ldr r3, =TabSin
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mov r4, #0 ; r4 = i
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mov r5, #0 ; r5 = ik
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mov r7, #0 ; r7 = 0
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mov r9, #0 ; r9 = résultat réel
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mov r10, #0 ; r10 = résultat imaginaire
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deb_somme ldrh r4, [r0, r3, LSL #1] ; r4 = x[i], LSL #1 pour faire ×2 car chaque valeur est sur 2 octets | format 4.12
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ldrsh r6, [r2, r5, LSL #1] ; r6 = cos ou sin [ik%N] | format 1.15
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mla r7, r4, r6, r7 ; r7 += x[i] * cos|sin[ik%n] | format 5.27
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deb_somme ldrh r6, [r0, r4, LSL #1] ; r6 = x[i], LSL #1 pour faire ×2 car chaque valeur est sur 2 octets | format 4.12
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ldrsh r7, [r2, r5, LSL #1] ; r7 = cos [ik%N] | format 1.15
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ldrsh r8, [r3, r5, LSL #1] ; r8 = sin [ik%N] | format 1.15
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mla r9, r6, r7, r9 ; r9 += x[i] * cos[ik%n] | format 5.27
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mla r10, r6, r8, r10 ; r10 += x[i] * sin[ik%n] | format 5.27
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add r5, r1 ; r5 = (i-1)k + k
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and r5, #N-1 ; r5 = r5 % 64
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add r3, #1 ; on incrémente i
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cmp r3, #N ;
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add r4, #1 ; on incrémente i
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cmp r4, #N ;
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bne deb_somme
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; etat des registres : r0 = x, r1 = k, r9 = Re, r10 = Im, autres libres
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smull r2, r0, r9, r9 ; (r2,r0) = Re(k)² | format : 10.54
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mov r0, r7
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pop {r4, r5, r6, r7}
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smlal r2, r0, r10, r10 ; (r2,r0) = Re(k)² + Im(k)² | format : 10.54
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pop {pc, r4, r5, r6, r7, r8, r9, r10}
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endp
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cal_comp proc ; r0 = x (signal), r1 = k, r2 = sin ou cos
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bx lr
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endp
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@ -4,6 +4,8 @@ extern short* TabSig[N];
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int resultats[N];
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int main(void) {
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resultats[5] = dft(TabSig, 5);
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for(int i = 1; i < N; ++i){
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resultats[i] = dft(TabSig, i);
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}
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while(1){}
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}
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